Leetcode——287. Find the Duplicate Number

题目原址

https://leetcode.com/problems/find-the-duplicate-number/description/

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.

Example1：

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

Example2：

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.

解题思路

给定一个数组，数组中包含n + 1个元素，数组的元素值范围为1 - n，因此，其中包含重复的元素，现在要求找到这个重复的元素并返回。

因为只有一个元素是重复的，所以这个题的思路就是：把每个元素的值 - 1作为新的数组元素的下标值，找到一个元素，就将新数组元素对应的值设置为 -1，遍历完整个数组，直到找到重复的元素为止。

AC代码

class Solution {
    public int findDuplicate(int[] nums) {
        int ret = 0;
        int[] num = new int[nums.length - 1];

        for(int i = 0; i < nums.length; i++) {
            if(!(num[nums[i] - 1] == -1))
                num[nums[i] - 1] = -1;
            else
                ret = nums[i];
        }
        return ret;       
    }
}