287. 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:

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.

思路1：二分法，count是nums里在head和mid之间的个数，如果重复数在head和mid之间，count>mid-head+1;否则在mid+1和tail之间。

public int findDuplicate(int[] nums) {        //binary search        int h = 1, t=nums.length-1;        while(h<t){            int count=0;            int mid=(h+t)/2;            for(int i=0; i<nums.length; i++){                if(nums[i]>=h&&nums[i]<=mid) count++;            }            if(count>(mid-h+1)) t=mid;            else h=mid+1;        }        return h;    }

思路2: 找环，当前值为下一个值的下标，形成一个新链表，如果两个数相同，则指向同一个下标，则形成环。转变成链表中找环的起点问题。

    public int findDuplicate(int[] nums) {        //find circle        int p1=0, p2=0;        do{            p1=nums[p1];            p2=nums[p2];            p2=nums[p2];        }        while(p1!=p2);        p1=0;        while(p1!=p2){            p1=nums[p1];            p2=nums[p2];        }        return p2;    }