题目描述

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.

样例

Example 1:

Input: [1,3,4,2,2]
Output: 2
Example 2:

Input: [3,1,3,4,2]
Output: 3
Note:

You must not modify the array (assume the array is read only).
You must use only constant, O(1) extra space.
Your runtime complexity should be less than O(n2).
There is only one duplicate number in the array, but it could be repeated more than once.

算法1

(二分) $O(nlogn)$

根据所有元素都在1->n之间，二分法计算小于等于mid的数量与mid比较，然后进行二分。

C++ 代码

class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        int left=1,right=nums.size()-1;
        while(left<right){//二分法判断重复元素在哪边
            int mid=left+(right-left)/2,cnt=0;
            for(auto num:nums){
                if(num<=mid)
                    cnt++;
            }
            if(cnt<=mid)
                left=mid+1;
            else
                right=mid;
        }
        return right;
    }
};

算法2

(暴力枚举) $O(n^2)$

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时间复杂度分析：blablabla

C++ 代码

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