二刷34

Author: diguage

docs/0000-01-modified-binary-search.adoc

@@ -1,5 +1,5 @@
 [#0000-01-modified-binary-search]
-= Modified Binary Search 改造过的二分搜索
+= Modified Binary Search 改进的二分搜索
 
 当你需要解决的问题的输入是排好序的数组，链表，或是排好序的矩阵，要求咱们寻找某些特定元素。这个时候的不二选择就是二分搜索。这种模式是一种超级牛的用二分来解决问题的方式。
 

docs/0034-find-first-and-last-position-of-element-in-sorted-array.adoc

@@ -1,39 +1,70 @@
 [#0034-find-first-and-last-position-of-element-in-sorted-array]
-= 34. Find First and Last Position of Element in Sorted Array
+= 34. 在排序数组中查找元素的第一个和最后一个位置
 
-{leetcode}/problems/find-first-and-last-position-of-element-in-sorted-array/[LeetCode - Find First and Last Position of Element in Sorted Array^]
+https://leetcode.cn/problems/find-first-and-last-position-of-element-in-sorted-array/[LeetCode - 34. 在排序数组中查找元素的第一个和最后一个位置 ^]
 
-Given an array of integers `nums` sorted in ascending order, find the starting and ending position of a given `target` value.
+给你一个按照非递减顺序排列的整数数组 `nums`，和一个目标值 `target`。请你找出给定目标值在数组中的开始位置和结束位置。
 
-Your algorithm's runtime complexity must be in the order of _O_(log _n_).
+如果数组中不存在目标值 `target`，返回 `[-1, -1]`。
 
-If the target is not found in the array, return `[-1, -1]`.
+你必须设计并实现时间复杂度为 stem:[log_2n] 的算法解决此问题。
 
-*Example 1:*
+*示例 1：*
 
 [subs="verbatim,quotes,macros"]
 ----
-*Input:* nums = [`5,7,7,8,8,10]`, target = 8
-*Output:* [3,4]
+输入：nums = [5,7,7,8,8,10], target = 8
+输出：[3,4]
 ----
 
-*Example 2:*
+*示例 2：*
 
 [subs="verbatim,quotes,macros"]
 ----
-*Input:* nums = [`5,7,7,8,8,10]`, target = 6
-*Output:* [-1,-1]
+输入：nums = [5,7,7,8,8,10], target = 6
+输出：[-1,-1]
 ----
 
+*示例 3：*
+
+[subs="verbatim,quotes,macros"]
+----
+输入：nums = [], target = 0
+输出：[-1,-1]
+----
+
+*提示：*
+
+* `+0 <= nums.length <= 10+`^`+5+`^
+* `-10`^`9`^`+<= nums[i] <= 10+`^`9`^
+* `nums` 是一个非递减数组
+* `-10`^`9`^`+<= target <= 10+`^`9`^
+
 == 思路分析
 
 可以直接使用二分查找，搜索两次。
 
 [[src-0034]]
+[tabs]
+====
+一刷::
++
+--
 [{java_src_attr}]
 ----
 include::{sourcedir}/_0034_FindFirstAndLastPositionOfElementInSortedArray.java[tag=answer]
 ----
+--
+
+二刷::
++
+--
+[{java_src_attr}]
+----
+include::{sourcedir}/_0034_FindFirstAndLastPositionOfElementInSortedArray_2.java[tag=answer]
+----
+--
+====
 
 == 参考资料
 

logbook/202503.adoc

@@ -15,6 +15,11 @@ endif::[]
 |{doc_base_url}/0206-reverse-linked-list.adoc[题解]
 |⭕️ 递归解法非常妙！传一个参数，`next` 又 `next` 比较麻烦，传两个参数比较简单。
 
+|{counter:codes2503}
+|{leetcode_base_url}/find-first-and-last-position-of-element-in-sorted-array/[34. Find First and Last Position of Element in Sorted Array^]
+|{doc_base_url}/0034-find-first-and-last-position-of-element-in-sorted-array.adoc[题解]
+|✅ 思考清楚确定边界时，中间指针的移动方向即可迎刃而解。
+
 
 |===
 

src/main/java/com/diguage/algo/leetcode/_0034_FindFirstAndLastPositionOfElementInSortedArray.java

@@ -31,7 +31,11 @@
  */
 public class _0034_FindFirstAndLastPositionOfElementInSortedArray {
   // tag::answer[]
-  public static int[] searchRange(int[] nums, int target) {
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2018-09-16 20:50 初次完成，2024-07-01 17:24:23 优化
+   */
+  public int[] searchRange(int[] nums, int target) {
     if (nums == null || nums.length == 0) {
       return new int[]{-1, -1};
     }

src/main/java/com/diguage/algo/leetcode/_0034_FindFirstAndLastPositionOfElementInSortedArray_2.java

@@ -0,0 +1,85 @@
+package com.diguage.algo.leetcode;
+
+
+public class _0034_FindFirstAndLastPositionOfElementInSortedArray_2 {
+  // tag::answer[]
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2025-03-04 21:02:59
+   */
+  public int[] searchRange(int[] nums, int target) {
+    int left = binarySearch(nums, target, true);
+    if (left == -1) {
+      return new int[]{-1, -1};
+    }
+    int right = binarySearch(nums, target, false);
+    return new int[]{left, right};
+  }
+
+  private int binarySearch(int[] nums, int target, boolean isLeft) {
+    int left = 0;
+    int right = nums.length - 1;
+    // 使用 result 变量，省去很多繁琐的判断
+    int result = -1;
+    while (left <= right) {
+      int mid = left + (right - left) / 2;
+      if (nums[mid] < target) {
+        left = mid + 1;
+      } else if (target < nums[mid]) {
+        right = mid - 1;
+      } else {
+        if (isLeft) {
+          // 注意：找左边界，要收缩右指针
+          right = mid - 1;
+        } else {
+          // 注意：找右边界，要搜索左指针
+          left = mid + 1;
+        }
+        result = mid;
+      }
+    }
+    return result;
+  }
+
+  // 下面是原始代码，上面是优化后的代码
+  private int binarySearchLeft(int[] nums, int target) {
+    int left = 0;
+    int right = nums.length - 1;
+    // 使用 result 变量，省去很多繁琐的判断
+    int result = -1;
+    while (left <= right) {
+      int mid = left + (right - left) / 2;
+      if (nums[mid] < target) {
+        left = mid + 1;
+      } else if (target < nums[mid]) {
+        right = mid - 1;
+      } else {
+        // 注意：找左边界，要收缩右指针
+        right = mid - 1;
+        result = mid;
+      }
+    }
+    return result;
+  }
+
+  private int binarySearchRight(int[] nums, int target) {
+    int left = 0;
+    int right = nums.length - 1;
+    // 使用 result 变量，省去很多繁琐的判断
+    int result = -1;
+    while (left <= right) {
+      int mid = left + (right - left) / 2;
+      if (nums[mid] < target) {
+        left = mid + 1;
+      } else if (target < nums[mid]) {
+        right = mid - 1;
+      } else {
+        // 注意：找右边界，要搜索左指针
+        left = mid + 1;
+        result = mid;
+      }
+    }
+    return result;
+  }
+  // end::answer[]
+}