{LeetCode} 457. Circular Array Loop

##此问题 Test Case并不完全，暂时AC。

You are given an array of positive and negative integers. If a number n at an index is positive, then move forward n steps. Conversely, if it's negative (-n), move backward n steps. Assume the first element of the array is forward next to the last element, and the last element is backward next to the first element. Determine if there is a loop in this array. A loop starts and ends at a particular index with more than 1 element along the loop. The loop must be "forward" or "backward'.

Example 1:Given the array [2, -1, 1, 2, 2], there is a loop, from index 0 -> 2 -> 3 -> 0.

Example 2:Given the array [-1, 2], there is no loop.

Note:The given array is guaranteed to contain no element "0".

Can you do it in O(n) time complexity and O(1) space complexity?

C++代码片：

class Solution {
public:
    bool circularArrayLoop(vector<int>& nums) {
        //vector<bool> check(nums.size(),false);
        for(int i=0; i<nums.size(); i++)
        {
            if(nums.at(i)==0) continue;
            int slow = i;
            int fast = i;
            cout<<move(nums,fast)<<endl;
            while(nums.at(i)*nums.at(slow)>0&&nums.at(i)*nums.at(fast)>0&&nums.at(i)*nums.at(move(nums,fast))>0)
            {
                slow = move(nums,slow);
                fast = move(nums,move(nums,fast));
                if(slow == fast)
                {
                    if(slow==move(nums,slow)) break;
                    else return true;
                }
            }
            slow = i;
            while(nums.at(i)*nums.at(slow)>0)
            {
                nums.at(slow) = 0;
                slow = move(nums,slow);
            }
        }
        return false;
    }
    
    int move(vector<int>& nums, int i)
    {
        int n = nums.size();
        return ((i+nums.at(i)>=0) ? ((i+nums.at(i))%n) : ((i+nums.at(i))%n)+n);
    }
};