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

桃扇骨 2022-06-04 10:26 202阅读 0赞

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);
        }
    };