64.Find the Duplicate Number（发现重复数字）-优快云博客

本文介绍了一种在不修改原始数组且使用常数级额外空间的情况下，查找包含n+1个整数的数组中重复数字的方法。通过设置快慢指针在循环链表中寻找相同的元素，最终找到重复的数字。

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Level：

Medium

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.

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.

思路分析：

和在一个循环链表中找两个相同元素的道理一样，设置一个快指针，一个慢指针，如果快指针和慢指针相同，则让快指针先走一步然后直到快指针的值等于慢指针的值，则找到答案。

代码：

public class Solution{
    public int findDuplicate(int []nums){
        int n=nums.length;
        int slow=0;
        int fast=1;
        while(nums[slow]!=nums[fast]){
            slow=(slow+1)%n;
            fast=(fast+2)%n;
            if(slow==fast){
                fast=(fast+1)%n;
            }
        }
        return nums[slow];
    }
}

转载于:https://www.cnblogs.com/yjxyy/p/11090462.html