35. Search Insert Position

问题：

Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.

You may assume no duplicates in the array.

Here are few examples.

[1,3,5,6]

, 5 → 2

[1,3,5,6]

, 2 → 1

[1,3,5,6]

, 7 → 4

[1,3,5,6]

, 0 → 0

思路：寻找目标数字在已经排好序的数组中的索引，若不存在则返回应该insert在什么地方，返回这个地方的索引。对于这种有序的数组问题，一般都要想到二分查找，或者二分插入

代码：

class Solution {
public:
int searchInsert(vector<int>& nums, int target) {
int low=0,high=nums.size()-1; //记录最初的low和high位置
while(low<=high)
{
int mid=low+(high-low)/2; //比较中间值

if(nums[mid]<target)
low=mid+1; //更新
else
high=mid-1; //更新
}
return low;
}
};