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.

Reference

 public int FindDuplicate(int[] nums) {
           int slow = 0;
            int fast = 0;
           
            while(true)
            {
                slow = nums[slow];
                fast = nums[nums[fast]];
                if (slow == fast) break;
            }
             int t=0;
            while(true)
            {
                 slow = nums[slow];
                 t = nums[t];
                if (slow == t) break;
            }
            return slow;
    }

 另一种解法是用binary search的方法，则方法为O(nlgn)

 public int FindDuplicate(int[] nums) {
          int left = 1;
          int n = nums.Count()-1;
          int right = nums.Count()-1;
          while(left< right)
          {
              int mid = left + (right - left)/2;
              int sumLeft = 0;
              for(int i = 0 ;i< nums.Count();i++)
              {
                if(nums[i]<=mid) sumLeft++;
              }
              if(sumLeft <= mid) left = mid+1;
              else right = mid;
          }
          return left;
    }