Hard 287题 Find the Duplicate Number

Question:

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:

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.

Solution:

This is a bad solution....but it works....

public class Solution {    public int findDuplicate(int[] nums) {        int n=nums.length;        int i=0;        int j=0;        if(n<=2)            return nums[0];        int ans=0;        for(i=0;i<=n-2;i++)        {            for(j=i+1;j<=n-1;j++)            {                if(nums[i]-nums[j]==0){                    ans=nums[i];break;}            }        }        return ans;    }}

快慢指针找环 参考了这个~http://www.jianshu.com/p/ce7f035daf74 写的好清楚

public class Solution {    public int findDuplicate(int[] nums) {        int fast=0;        int slow=0;        if(nums == null || nums.length < 2) {            return -1;        }        while(true)        {            slow=nums[slow];            fast=nums[nums[fast]];            if(slow==fast)                break;        }        fast=0;        while(true)        {            slow=nums[slow];            fast=nums[fast];            if(slow==fast)                break;        }        return fast;            }}

 还可以用二分法做。。据说也很棒！我觉得有点难度，巧妙用了nums[i]和i，抽屉原理

int n=nums.length;      int low=0;      int high=n-1;      while(low<high)      {          int mid=(high-low)/2+low;          int c=count(nums,mid);           if(c<=mid) {                  low = mid+1;              } else {                  high = mid;              }        }      return low;    }    public int count(int nums[], int mid)    {        int ans=0;        for(int i=0;i<=nums.length-1;i++)        {            if(nums[i]<=mid)                ans++;        }        return  ans;