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).

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

解决方法，根据数值的范围和数组的大小，我们可以使用下标来做文章，如果一个位置被访问了两个，那么

将说明该位置的元素是重复的，那么我们怎么知道这件事呢，可以给位置的数值做文章，下面是相关的代码

  

int findDuplicate(vector<int>& nums) {

    if(nums.size() ==0) return -1;

    int pos =0;

    size_t len = nums.size();

    for(int i =0;i < len;++i){

        pos = abs(nums[i]);

        if(nums[pos] >0)

            nums[pos] = -nums[pos];

        elsebreak;

    }

    return pos;

}