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

本题给定的条件是数组，数组里面有n+1个整数，其中包括1～n个数，其中有一个数是重复的，求那个重复的数。

 如图所示：

此题分为两个steps

1.第一次相遇的时候，2m = m+c*r-------->m=c*r@;

n+a=m@

结合起来就是n+a=c*r------>n=c*r-a------>n=(c-1)*r+r-a;

2.通过之前的公式可以知道，把其中一个指针重置为起点后，而另一个指针不变，每次移动一步，那么下一次相遇的地点就是入口处，而本题数组循环的入口处是肯定为重复数字的，因此问题解决了，代码如下：

public class Solution {
    public int findDuplicate(int[] nums) {
        //we set the slow and fast pointer to the index of 0,so that they could reach in the same loop
        int slow = nums[0];
        int fast = nums[nums[0]];
        //find the meeting point in the loop
        while(slow!=fast){
            slow = nums[slow];
            fast = nums[nums[fast]];
        }
        //find the entry loop point
        fast = 0;
        while(slow!=fast){
            slow = nums[slow];
            fast = nums[fast];
        }
        return slow; 
    }
}
//the run time could be less than O(n);the space complexity coud be O(1);

 此题也是说明了一点，即起点设为哪一点都可以，最终的结果都不受影响

public class Solution {
    public int findDuplicate(int[] nums) {
        int low = 1;
        int high = nums.length-1;
        while( low < high ) {
            int count = 0;
            int mid = low + ( high - low ) / 2;
            for(int i=0;i<nums.length;i++){
                if(nums[i]<=mid) count++;
            }
            if(count<=mid) low = mid+1;
            else high = mid;
        }
        return low;
    }
}
// the run time complexity could be O(nlogn) ,the space complexity could be O(1);