457. Circular Array Loop

問題描述：

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?

 

解題思路：

首先咱們要明確怎樣算是一個環：

　　1. 起始座標和結束座標爲同一座標

　　2. 環中要有多於一個的元素

　　3. 環須要是單向的。即要麼只向前，要麼只向後。

首先根據題意咱們能夠構造一個輔助方法：getNextIdx，找該點下個點。

這裏須要注意的是！

數組中存在的環的起始點不定，因此咱們要對每個點爲起始點存在的環來進行判斷。

 

代碼：

class Solution {
public:
    bool circularArrayLoop(vector<int>& nums) {
        if(nums.size() == 0) return false;
        for(int i = 0; i < nums.size(); i++){
            if(isLoop(nums, i)) return true;
        }
        return false;
    }
    int getNextIdx(vector<int>& nums, int cur){
        int len = nums.size();
        cur = cur + nums[cur];
        if(cur > 0) cur = cur % len;
        else cur = len - abs(cur)%len;
        return cur;
    }
    bool isLoop(vector<int> nums, int i){
        int slow = i, fast = i;
        int len = nums.size();
        do{
            slow = getNextIdx(nums, slow);
            fast = getNextIdx(nums, fast);
            fast = getNextIdx(nums, fast);
        }while(slow != fast);
        int nxt = getNextIdx(nums, slow);
        if(nxt == slow) return false;
        int direction = nums[fast] / abs(nums[fast]);
        int start = fast;
        do{
            if(nums[start] * direction < 0) return false;
            start = getNextIdx(nums, start);
        }while(start != fast);
        
        return true;
    }
};