Find the Duplicate Number -- LeetCode

最新推荐文章于 2018-11-30 20:22:09 发布

转载最新推荐文章于 2018-11-30 20:22:09 发布 · 55 阅读

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原文链接：http://www.cnblogs.com/fenshen371/p/5162178.html

本文介绍了一种在不修改原始数组且仅使用常数级额外空间的情况下查找数组中重复数字的方法。该方法利用快慢指针技巧，确保时间复杂度低于O(n²)，适用于仅有一个重复数字的情况。

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:

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：与这一道题的思路相同传送门

 1 class Solution {
 2 public:
 3     int findDuplicate(vector<int>& nums) {
 4         int slow = nums[0], fast = nums[nums[0]];
 5         while (slow != fast)
 6         {
 7             slow = nums[slow];
 8             fast = nums[nums[fast]];
 9         }
10         fast = 0;
11         while (slow != fast)
12         {
13             slow = nums[slow];
14             fast = nums[fast];
15         }
16         return slow;
17     }
18 };

转载于:https://www.cnblogs.com/fenshen371/p/5162178.html