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Add leetcode exercises
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Diff for: ‎README.md

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Diff for: ‎algorithms/001-two-sum.ipynb

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{
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"cells": [
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{
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"cell_type": "markdown",
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"id": "common-configuration",
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"metadata": {},
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"source": [
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"给定一个整数数组 nums 和一个目标值 target，请你在该数组中找出和为目标值的那 两个 整数，并返回他们的数组下标。\n",
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"你可以假设每种输入只会对应一个答案。但是，数组中同一个元素不能使用两遍。\n",
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"\n",
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"#### 示例:\n",
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"- 给定 `nums = [2, 7, 11, 15]`, `target = 9`\n",
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"- 因为 `nums[0] + nums[1] = 2 + 7 = 9`\n",
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"- 所以返回 `[0, 1]`"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 4,
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"id": "black-discovery",
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"metadata": {},
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"outputs": [],
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"source": [
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"def solve(nums, target):\n",
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" candidates = {}\n",
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" for i, n in enumerate(nums):\n",
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" candidates[target - n] = i\n",
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" for i, n in enumerate(nums):\n",
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" if n in candidates and i != candidates[n]:\n",
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" return [i, candidates[n]]\n",
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" return None"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 5,
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"id": "refined-palestinian",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"[0, 1]"
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]
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},
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"execution_count": 5,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"solve([2, 7, 11, 15], 9)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 6,
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"id": "early-reader",
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"[1, 3]\n",
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"[1, 2]\n",
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"[0, 3]\n",
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"[2, 4]\n"
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]
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}
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],
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"source": [
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"print(solve([11, 7, 15, 2], 9))\n",
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"print(solve([3, 2, 4], 6))\n",
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"print(solve([0, 4, 3, 0], 0))\n",
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"print(solve([-1, -2, -3, -4, -5], -8))"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"id": "little-demand",
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"metadata": {},
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"outputs": [],
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"source": []
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.8.6"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 5
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}

Diff for: ‎algorithms/001-two-sum.py

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"""
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给定一个整数数组 nums 和一个目标值 target，请你在该数组中找出和为目标值的那 两个 整数，并返回他们的数组下标。
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你可以假设每种输入只会对应一个答案。但是，数组中同一个元素不能使用两遍。
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#### 示例:
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给定 nums = [2, 7, 11, 15], target = 9
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因为 nums[0] + nums[1] = 2 + 7 = 9
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所以返回 [0, 1]
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"""
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def two_sum(nums, target: int):
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candidates = {}
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for i, n in enumerate(nums):
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# if target >= n:
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candidates[target - n] = i
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for i, n in enumerate(nums):
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if n in candidates and i != candidates[n]:
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return [i, candidates[n]]
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if __name__ == '__main__':
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print(two_sum([2, 7, 11, 15], 9))
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print(two_sum([11, 7, 15, 2], 9))
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print(two_sum([3, 2, 4], 6))
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print(two_sum([0, 4, 3, 0], 0))
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print(two_sum([-1, -2, -3, -4, -5], -8))

Diff for: ‎algorithms/002-add-two-numbers.ipynb

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{
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"cells": [
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{
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"cell_type": "markdown",
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"id": "stainless-cross",
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"metadata": {},
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"source": [
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"### [两个链表生成相加链表](https://leetcode-cn.com/problems/add-two-numbers/)\n",
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"\n",
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"给你两个**非空**的链表，表示两个非负的整数。它们每位数字都是按照**逆序**的方式存储的，并且每个节点只能存储**一位**数字。\n",
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"\n",
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"请你将两个数相加，并以相同形式返回一个表示和的链表。\n",
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"\n",
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"你可以假设除了数字`0`之外，这两个数都不会以`0`开头。\n",
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"\n",
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"#### 示例 1：\n",
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"```\n",
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"输入：l1 = [2,4,3], l2 = [5,6,4]\n",
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"输出：[7,0,8]\n",
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"解释：342 + 465 = 807.\n",
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"```\n",
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"\n",
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"#### 示例 2：\n",
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"```\n",
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"输入：l1 = [0], l2 = [0]\n",
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"输出：[0]\n",
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"```\n",
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"\n",
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"#### 示例 3：\n",
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"```\n",
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"输入：l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]\n",
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"输出：[8,9,9,9,0,0,0,1]\n",
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"```\n",
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"#### 提示：\n",
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"- 每个链表中的节点数在范围`[1, 100]`内\n",
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"- `0 <= Node.val <= 9`\n",
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"- 题目数据保证列表表示的数字不含前导零"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 3,
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"id": "tested-consideration",
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"metadata": {},
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"outputs": [],
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"source": [
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"class ListNode:\n",
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" def __init__(self, val=0, next=None):\n",
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" self.val = val\n",
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" self.next = next\n",
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"\n",
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"class Solution:\n",
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" def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:\n",
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" h1, h2, ans = l1, l2, ListNode()\n",
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" carry, curr = 0, ans\n",
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" while h1 or h2:\n",
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" h1val = h1.val if h1 else 0\n",
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" h2val = h2.val if h2 else 0\n",
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" value = h1val + h2val + carry\n",
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" if value >= 10:\n",
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" carry = 1\n",
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" value = value % 10\n",
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" else:\n",
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" carry = 0\n",
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" curr.next = ListNode(value)\n",
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" if h1 is not None:\n",
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" h1 = h1.next\n",
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" if h2 is not None:\n",
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" h2 = h2.next\n",
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" curr = curr.next\n",
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" \n",
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" if carry > 0:\n",
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" curr.next = ListNode(carry)\n",
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" return ans.next"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"id": "joint-retro",
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"metadata": {},
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"outputs": [],
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"source": []
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.8.6"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 5
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}

Diff for: ‎algorithms/002-add-two-numbers.py

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"""
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两数相加
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给出两个 非空 的链表用来表示两个非负的整数。其中，它们各自的位数是按照 逆序 的方式存储的，并且它们的每个节点只能存储 一位 数字。
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如果，我们将这两个数相加起来，则会返回一个新的链表来表示它们的和。
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您可以假设除了数字 0 之外，这两个数都不会以 0 开头。
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示例：
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输入：(2 -> 4 -> 3) + (5 -> 6 -> 4)
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输出：7 -> 0 -> 8
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原因：342 + 465 = 807
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"""
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class ListNode:
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def __init__(self, x):
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self.val = x
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self.next = None
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# def add_two_numbers(l1: ListNode, l2: ListNode) -> ListNode:
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# p, q = l1, l2
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# curr = head = ListNode(None)
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# carry = 0
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# while p is not None and q is not None:
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# val = p.val + q.val + carry
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# if val >= 10:
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# val = val % 10
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# carry = 1
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# else:
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# carry = 0
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#
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# curr.next = ListNode(val)
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# curr = curr.next
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# p, q = p.next, q.next
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#
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# if carry == 0:
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# if p is not None:
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# curr.next = p
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# elif q is not None:
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# curr.next = q
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# else:
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# while p is not None:
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# val = p.val + carry
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# if val >= 10:
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# val = val % 10
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# curr.next = ListNode(val)
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# curr = curr.next
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# else:
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# curr.next = ListNode(val)
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# curr = curr.next
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# curr.next = p.next
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# carry = 0
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# break
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# p = p.next
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#
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# while q is not None:
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# val = q.val + carry
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# if val >= 10:
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# val = val % 10
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# curr.next = ListNode(val)
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# curr = curr.next
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# else:
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# curr.next = ListNode(val)
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# curr = curr.next
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# curr.next = q.next
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# carry = 0
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# break
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# q = q.next
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#
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# if carry == 1:
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# curr.next = ListNode(1)
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#
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# return head.next
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def add_two_numbers(l1: ListNode, l2: ListNode) -> ListNode:
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p, q = l1, l2
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curr = head = ListNode(None)
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carry = 0
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while p is not None or q is not None:
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pval = p.val if p is not None else 0
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qval = q.val if q is not None else 0
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val = pval + qval + carry
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if val >= 10:
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val = val % 10
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carry = 1
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else:
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carry = 0
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curr.next = ListNode(val)
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curr = curr.next
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if p is not None:
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p = p.next
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if q is not None:
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q = q.next
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if carry == 1:
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curr.next = ListNode(1)
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return head.next
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if __name__ == '__main__':
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l1 = ListNode(9)
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l1.next = ListNode(8)
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l2 = ListNode(1)
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print(add_two_numbers(l1, l2))
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