标签：class font note this min 存在 strong 解题思路 int
问题描述：
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?
解题思路：
首先我们要明确怎样算是一个环：
1. 起始坐标和结束坐标为同一坐标
2. 环中要有多于一个的元素
3. 环需要是单向的。即要么只向前，要么只向后。
首先根据题意我们可以构造一个辅助方法：getNextIdx，找该点下个点。
这里需要注意的是！
数组中存在的环的起始点不定，所以我们要对每一个点为起始点存在的环来进行判断。
代码：
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; } };
标签：class font note this min 存在 strong 解题思路 int
原文地址：https://www.cnblogs.com/yaoyudadudu/p/9393758.html
