Sword refers to offer 14. Do not modify the array to find repeated numbers (two points)

Topic

answer

1. If additional arrays can be used, we can open a new array to count the number of occurrences of each number.

2. We can use dichotomy to do this question, because n+1 numbers in 1-n must have two numbers that are repeated, so every time we divide by the middle value, then there must be a number of intervals on the left and right. Is greater than its l-mid+1, this greater than is the repeated number

Code

class Solution {
    
    
public:
    int duplicateInArray(vector<int>& nums) {
    
    
        int l=1,r=nums.size()-1;
        while(l<r){
    
    
            int mid=l+r>>1;
            int s=0;
            for(auto x:nums) {
    
    
                if(x>=l&&x<=mid){
    
    
                    s++;
                }
            }
            if(s>mid-l+ 1) r = mid;
            else l = mid + 1;
        }
        return l;
    }
};