Find the Duplicate Number

Find the Duplicate Number

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.

分析1

比较容易想到的是O(nlogn)的算法，根据题意，所有的数字小于等于n，大于等于1，可以通过二分查找的方法，首先查找小于中间数字的数量，如果重复的数字比中间数字小，那么小于中间数字的数量肯定大于中间数字，反之则重复的数字比中间的数字大

代码1

class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        int high = nums.size()-1;
        int low = 1;

        while (low < high){
            int mid = (high+low)/2;
            int count = 0;
            for (int num: nums){
                if (num <= mid) count++;
            }

            if (count <= mid) low = mid+1;
            else high = mid;
        }

        return low;
    }
};

分析2

题目中最关键的信息是，所有的数字介于1到n之间，并且总数为n+1，那么每一个位置存储的数字就可以看作是数组中的位置，那么整个数组就可以看成是一个链表，头节点是nums[0]，如果存在重复的数字，那就是有不同的节点指向同一个位置，即存在环，所以整个问题就转化成了单向链表求环的问题，并且，环相交的点就是重复的数字。

代码2

class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        int fast = nums[nums[0]];
        int low = nums[0];

        while (low != fast){
            low = nums[low];
            fast = nums[nums[fast]];
        }

        low = 0;
        while (low != fast){
            low = nums[low];
            fast = nums[fast];
        }

        return low;
    }
};