[LeetCode] Search for a Range

最新推荐文章于 2025-09-09 19:16:33 发布

weixin_33686714

最新推荐文章于 2025-09-09 19:16:33 发布

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文章标签：开发工具

原文链接：https://yq.aliyun.com/articles/331400

本文介绍了一种通过两次二分搜索来寻找目标元素左右边界的算法实现，详细阐述了如何通过巧妙的技巧来优化搜索过程。

The idea is to search for the left and right boundaries of target via two binary searches. Well, some tricks may be needed. Take a look at this link :-)

The code is rewritten as follows.

 1 class Solution {
 2 public:
 3     vector<int> searchRange(vector<int>& nums, int target) {
 4         int l = left(nums, target);
 5         if (l == -1) return {-1, -1};
 6         return {l, right(nums, target)};
 7     }
 8 private:
 9     int left(vector<int>& nums, int target) {
10         int n = nums.size(), l = 0, r = n - 1;
11         while (l < r) {
12             int m = l + ((r - l) >> 1);
13             if (nums[m] < target) l = m + 1;
14             else r = m;
15         }
16         return nums[l] == target ? l : -1;
17     }
18     int right(vector<int>& nums, int target) {
19         int n = nums.size(), l = 0, r = n - 1;
20         while (l < r) {
21             int m = l + ((r - l + 1) >> 1); 
22             if (nums[m] > target) r = m - 1;
23             else l = m;
24         }
25         return r;
26     }
27 };