<LeetCode OJ> 287. Find the Duplicate Number

287. Find the Duplicate Number

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Total Accepted: 18097 Total
Submissions: 48596 Difficulty: Hard

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:

You must not modify the array (assume the array is read only).
You must use only constant, O(1) extra space.
Your runtime complexity should be less than

O(n2)

.
There is only one duplicate number in the array, but it could be repeated more than once.

Credits:

Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

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分析1：

本题提示为双指针问题，不知道怎么做！！！

本人的，不符合要求的做法，有空间复杂度！

class Solution {
public:
    int findDuplicate(vector<int>& nums) {
       unordered_set<int> uset;
       for (int i =0;i < nums.size(); i++) 
       {
            if(uset.find(nums[i])!=uset.end())
                return nums[i];
            else
                uset.insert(nums[i]);
        }
    }
};

以下为鉴赏别人的分析：

1). 使用快慢指针法，若链表中有环，可以得到两指针的交点M
2). 记链表的头节点为H，环的起点为E

2.1) L1为H到E的距离
2.2) L2为从E出发，首次到达M时的路程
2.3) C为环的周长
2.4) n为快慢指针首次相遇时，快指针在环中绕行的次数

根据L1,L2和C的定义，我们可以得到：
慢指针行进的距离为L1 + L2
快指针行进的距离为L1 + L2 + n * C

由于快慢指针行进的距离有2倍关系，因此：
2 * (L1+L2) = L1 + L2 + n * C => L1 + L2 = n * C => L1 = (n - 1)* C + (C - L2)
可以推出H到E的距离 = 从M出发绕环到达E时的路程
因此，当快慢指针在环中相遇时，我们再令一个慢指针从头节点出发
接下来当两个慢指针相遇时，即为E所在的位置

参考代码为：

class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        int slow = nums[0], fast = nums[nums[0]];
        while(slow != fast) {
            slow = nums[slow];
            fast = nums[nums[fast]];
        }
        
        slow = 0;
        while(slow != fast) {
            slow = nums[slow];
            fast = nums[fast];
        }
        
        return slow;
    }
};

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原文地址：http://blog.csdn.net/ebowtang/article/details/50569543

原作者博客：http://blog.csdn.net/ebowtang