leetcode 457. Circular Array Loop 数组环+数组中寻找环的存在+双指针+循环判断

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?

题意很简单，就是一个循环的判断，这道题可以使用双指针来判断，要注意的是双指针的移动要注意保持方向一致

所以在while的地方我们要求当前的和fast和fast的方向是一致的，这样才能全部AC

代码如下：

#include <iostream>
#include <vector>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>
#include <functional>
#include <bitset>
#include <numeric>
#include <cmath>
#include <regex>

using namespace std;


class Solution 
{
public:
    bool circularArrayLoop(vector<int>& a) 
    {
        for (int i = 0; i < a.size(); i++)
        {
            if (a[i] == 0)
                continue;
            int slow = i, fast = getNextIndex(i, a);
            while (a[i] * a[fast] > 0 && a[i] * a[getNextIndex(fast, a)] > 0)
            {
                if (slow == fast)
                {
                    if (slow == getNextIndex(slow, a))
                        break;
                    else
                        return true;
                }
                else
                {
                    slow = getNextIndex(slow, a);
                    fast = getNextIndex(getNextIndex(fast, a), a);
                }
            }
        }
        return false;
    }

    int getNextIndex(int i, vector<int>& a)
    {
        int size = a.size();
        if (i + a[i] > 0)
            return (i + a[i]) % size;
        else
            return ((i + a[i]) % size + size);
    }
};