给定一个排序数组和一个目标值，在数组中找到目标值， 并返回其索引。如果目标值不存在于数组中， 返回它将会被按顺序插入的位置。

请必须使用时间复杂度为 $O(logn)$的算法。

示例 1:
输入: nums = [1,3,5,6], target = 5
输出: 2

示例 2:
输入: nums = [1,3,5,6], target = 2
输出: 1

示例 3:
输入: nums = [1,3,5,6], target = 7
输出: 4

提示:

• 1 <= nums.length <= $10^4$
• $-10^4$ <= nums[i] <= $10^4$
• nums 为 无重复元素 的 升序 排列数组
• $-10^4$ <= target <= $10^4$

题解1 二分查找

class Solution {
public:int searchInsert(vector<int>& nums, int target) {int len = nums.size();int left = 0;int right = len-1;int pos = 0;while(left <= right){int mid = (left+right) >> 1;if(nums[mid] == target) return mid;else if(nums[mid] < target){left = mid+1;// left就是升序情况下 应该插入的位置pos = left;}else{right = mid-1;}}return pos;}
};

防越界写法

class Solution {
public:int searchInsert(vector<int>& nums, int target) {int n = nums.size();int left = 0, right = n - 1, ans = n;while (left <= right) {int mid = ((right - left) >> 1) + left;if (target <= nums[mid]) {// mid = left + (difference)>>1 (Key: 找到第一个下标，对应值 >= target)ans = mid;right = mid - 1;} else {left = mid + 1;}}return ans;}
};

题解2 STL大法

class Solution {
public:int searchInsert(vector<int>& nums, int target) {auto it = find(nums.begin(),nums.end(),target);if(it!=nums.end()){return it-nums.begin();}auto first =lower_bound(nums.begin(), nums.end(), target);return first-nums.begin();}
};

两行

class Solution {
public:int searchInsert(vector<int>& nums, int target) {auto st = nums.cbegin(), ed = nums.cend();return lower_bound(st, ed, target) - st;}
};