一、朴素二分
. - 力扣(LeetCode). - 备战技术面试?力扣提供海量技术面试资源,帮助你高效提升编程技能,轻松拿下世界 IT 名企 Dream Offer。
https://leetcode.cn/problems/binary-search/description/
int left = 0, right = nums.size();
int mid = left + ((right - left) >> 1); // 防止溢出
if (nums[mid] == target) {
} else if (nums[mid] > target) {
二、整数二分
. - 力扣(LeetCode). - 备战技术面试?力扣提供海量技术面试资源,帮助你高效提升编程技能,轻松拿下世界 IT 名企 Dream Offer。
https://leetcode.cn/problems/find-first-and-last-position-of-element-in-sorted-array/description/
vector<int> searchRange(vector<int>& nums, int target) {
int l = 0, r = nums.size() - 1;
vector<int> ans = {-1, -1};
if (nums.size() == 0) return ans;
int mid = l + (r - l) / 2;
if (nums[mid] >= target) r = mid;
if (nums[l] == target) ans[0] = l;
int mid = l + (r - l + 1) / 2;
if (nums[mid] <= target) l = mid;