LeetCode：287. Find the Duplicate Number

一、问题描述

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.

Example 1:

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

Example 2:

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

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.

二、解题思路

使用set(因为涉及到查找，所以使用了底层是哈希表的unordered_set)，遍历一遍数组，如果这个值在set中存在，直接return；不存在的话，添加到set中

三、代码实现（时间复杂度O(N)）

#include<unordered_set>
class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        unordered_set<int> result;
        for(int num:nums){
            if(result.find(num) != result.end()){
                return num;
            }else {
                result.insert(num);
            }
        }
        return -1;
    }
};

说明其它的几种解题方法：

①排序后，判断当前元素和后一个是否相等，相等的话就找到了。但是这种方法的时间复杂度是：O(NlogN)

来源：https://blog.csdn.net/ixiaowei1993/article/details/100777324