287. Find the Duplicate Number

Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.

Note:

  1. You must not modify the array (assume the array is read only).
  2. You must use only constant, O(1) extra space.
  3. Your runtime complexity should be less than O(n2).

  1. There is only one duplicate number in the array, but it could be repeated more than once.
【思路】二分查找,小于mid的个数多于mid。则在左边,小于mid的个数少于mid,则在右边。

class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        
        int left = 0;
        int right = nums.size()-1;
        while(left < right){
            int mid = left + (right - left)  /2;
            int cnt1 = 0;
            for(int i = 0; i < nums.size(); i++){
                if(nums[i]<= mid)
                 cnt1++;
            }
            if(cnt1<=mid)
              left = mid +1;
           else
              right = mid;
        }
        return left;
    }
};


你可能感兴趣的:(287. Find the Duplicate Number)