README.md

560. 和为 K 的子数组

English Version

题目描述

给你一个整数数组 nums 和一个整数 k ，请你统计并返回 该数组中和为 k 的连续子数组的个数 。

子数组是数组中元素的连续非空序列。

 

示例 1：

输入：nums = [1,1,1], k = 2
输出：2

示例 2：

输入：nums = [1,2,3], k = 3
输出：2

 

提示：

• 1 <= nums.length <= 2 * 104
• -1000 <= nums[i] <= 1000
• -107 <= k <= 107

解法

Python3

class Solution:
    def subarraySum(self, nums: List[int], k: int) -> int:
        counter = Counter({0: 1})
        ans = s = 0
        for num in nums:
            s += num
            ans += counter[s - k]
            counter[s] += 1
        return ans

Java

class Solution {
    public int subarraySum(int[] nums, int k) {
        Map<Integer, Integer> counter = new HashMap<>();
        counter.put(0, 1);
        int ans = 0, s = 0;
        for (int num : nums) {
            s += num;
            ans += counter.getOrDefault(s - k, 0);
            counter.put(s, counter.getOrDefault(s, 0) + 1);
        }
        return ans;
    }
}

TypeScript

function subarraySum(nums: number[], k: number): number {
    let ans = 0,
        s = 0;
    const counter = new Map();
    counter.set(0, 1);
    for (const num of nums) {
        s += num;
        ans += counter.get(s - k) || 0;
        counter.set(s, (counter.get(s) || 0) + 1);
    }
    return ans;
}

C++

class Solution {
public:
    int subarraySum(vector<int>& nums, int k) {
        unordered_map<int, int> counter;
        counter[0] = 1;
        int ans = 0, s = 0;
        for (int& num : nums) {
            s += num;
            ans += counter[s - k];
            ++counter[s];
        }
        return ans;
    }
};

Go

func subarraySum(nums []int, k int) int {
	counter := map[int]int{0: 1}
	ans, s := 0, 0
	for _, num := range nums {
		s += num
		ans += counter[s-k]
		counter[s]++
	}
	return ans
}

Rust

impl Solution {
    pub fn subarray_sum(mut nums: Vec<i32>, k: i32) -> i32 {
        let n = nums.len();
        let mut count = 0;
        for i in 0..n {
            let num = nums[i];
            if num == k {
                count += 1;
            }
            for j in 0..i {
                nums[j] += num;
                if nums[j] == k {
                    count += 1;
                }
            }
        }
        count
    }
}

use std::collections::HashMap;

impl Solution {
    pub fn subarray_sum(nums: Vec<i32>, k: i32) -> i32 {
        let mut res = 0;
        let mut sum = 0;
        let mut map = HashMap::new();
        map.insert(0, 1);
        nums.iter().for_each(|num| {
            sum += num;
            res += map.get(&(sum - k)).unwrap_or(&0);
            map.insert(sum, map.get(&sum).unwrap_or(&0) + 1);
        });
        res
    }
}

...