[Leetcode] Find the Duplicate Number, Solution

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.

[Thoughts]

这题想清楚了就不难，想不清楚就麻烦。

假设数组A长度是n, 里面存储1到n的整数，那么很清楚，咱们能够在按照A[i] = i+1，进行排序。可是如今有n+1个整数，并且至少有一个数字存在冗余。若是咱们仍然按照A[i] = i+1来排序数组的话，那么当发现A[i]上已是i+1的值时，说明咱们已经找到了冗余数了。

举个例子，

简单的说，就是遍历数组的同时，按照A[i]应该放到A[A[i]]原则，进行swap，第一个没法swap的数字就是所求。

[Code]

1:  class Solution {  
2:  public:  
3:    int findDuplicate(vector<int>& nums) {  
4:      int length = nums.size();  
5:      for(int i =0; i< length; i++) {  
6:        if(nums[i] == i+1) {  
7:          continue;  
8:        }  
9:        int oldIndex = i;  
10:        int newIndex = nums[i]-1;  
11:        while(nums[oldIndex] != oldIndex +1 ) {  
12:          if(nums[oldIndex] == nums[newIndex] ) {  
13:            return nums[oldIndex];  
14:          }  
15:          int temp = nums[newIndex];  
16:          nums[newIndex] = nums[oldIndex];  
17:          nums[oldIndex] = temp;  
18:          newIndex = nums[oldIndex] -1;  
19:        }  
20:      }  
21:    }  
22:  };  

github: https://github.com/codingtmd/leetcode/blob/master/src/Find_the_Duplicate_Number.cpp