[C++知识库]力扣-剑指 Offer 53 - II. 0~n-1中缺失的数字（C++）- 二分法+有价值

题目链接：https://leetcode-cn.com/problems/que-shi-de-shu-zi-lcof/
题目如下：

思路：

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        int size=nums.size();
        if(size==0) return 0;

        int low=0,high=size-1,mid;

        while(low<high){
            mid=(low+high)/2;
            if(nums[mid]!=mid) high=mid;  //[L,R]->[L,M]
            else low=mid+1;   //nums[mid]==mid;[L,R]->[M+1,R]
        }
        
        if(nums[high]==high) high++;//整个序列是从0~n-1，丢失的数字为n

        return high;
    }
};