Leetcode 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.

最容易想到的思路是新开一个长度为n的全零list p[1~n]。依次从nums里读出数据，假设读出的是4， 就将p[4]从零改成1。如果发现已经是1了，那么这个4就已经出现过了，所以他就是重复的那个数。这个解法的时间复杂度是O(N)。但是由于本题要求空间复杂度是O(1)。所以不能用。

可以用二分法，low = 1, high = n, mid = (left + right)//2，如果<=mid 的元素个数 > mid，那么重复的数字一定在[1, mid]区间内。反之，则一定在[mid+1, high]里面。注意红色的>mid不能是>=。

例如 【1，2, 2， 3(mid)， 4， 5， 6】，【1，2, 3， 3(mid)， 4， 5】

 class Solution(object):
     def findDuplicate(self, nums):
         """
         :type nums: List[int]
         :rtype: int
         """
         left = 1
         right = len(nums) - 1

         while left < right:
             mid = (left + right)//2
             count = 0
             for x in nums:
                 if x <= mid:
                     count += 1

             if count > mid:
                 right = mid
             else:
                 left = mid + 1

         return left

第二种解法来自：http://www.cnblogs.com/grandyang/p/4843654.html

使用类似Linked List Cycle II的思路。

 def findDuplicate(self, nums):
         """
         :type nums: List[int]
         :rtype: int
         """
         slow = 0
         fast = 0
         t = 0
         while True:
             slow = nums[slow]
             fast = nums[nums[fast]]
             if slow == fast:
                 break
         while True:
             slow = nums[slow]
             t = nums[t]
             if slow == t:
                 break
         return slow