Find the Duplicate Number

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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).
4. There is only one duplicate number in the array, but it could be repeated more than once.

这里有个条件，就是数组中的数的范围是[1,n];

我们可以尝试用二分法，第一次取(1+n)/2，如果在整个数组中小于等于(1+n)/2的数的个数超过了(1+n)/2，那么重复的数一定是[1,(1+n)/2)区间的数。

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

Find the Duplicate Number

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原文地址：http://www.cnblogs.com/zengzy/p/5003724.html