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 
    class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        int min=1,max=nums.size();
        while(min<=max)
        {
            int mid=(min+max)/2;
            int cnt=0;
            for(int i=0;i<nums.size();i++)
                if(nums[i]<=mid)
                    cnt++;
            if(cnt>mid)
                max=mid-1;
            else
                min=mid+1;
        }
        return min;
    }
};

    　　二分查找的思路，如何1到n/2有重复的，那么1到n/2的数字个数一定大于n/2这个数本身。下面是网上别人的思路，看不懂。

    class Solution {
public:
int findDuplicate(vector<int>& nums) {
    int slow = 0;
    int fast = 0;
    int finder = 0;

    while (true)
    {
        slow = nums[slow];
        fast = nums[nums[fast]];

        if (slow == fast)
            break;
    }
    while (true)
    {
        slow = nums[slow];
        finder = nums[finder];
        if (slow == finder)
            return slow;
    }
}

    　　如果第一个数为0的话，不就死循环了吗？看来是特例特思。