解法一：二分查找法
class Solution {public int searchInsert(int[] nums, int target) {int n = nums.length;int left = 0, right = n - 1, location = n;while(left<=right){int mid = (left+right)/2;if(nums[mid]>=target){location = mid;right = mid-1;}else{left = mid+1;}}return location;}
}
注意：
location初始化为n（指针指向最右边，因为location用mid记录都是记录小于某个数应该在的位置，无法找到位置n），当nums[mid]>=target时更新为mid- 循环终止条件为
left<=right - 用
mid更新时，要记得+1 -1 - 时间复杂度为 O(log n) 的算法 -> 二分查找
int mid = (left+right)/2 每次都取中间或者中间左边的那个数int mid = (left+right+1)/2 每次都取中间或者中间右边的那个数
