https://leetcode.com/problems/circular-array-loop/
题目蛮难的,有一些坑。
前后两个指针追赶找环的方法,基本可以归结为一种定式。可以多总结。
package com.company; class Solution { // 参考了这里的解法 // https://discuss.leetcode.com/topic/66894/java-slow-fast-pointer-solution public boolean circularArrayLoop(int[] nums) { for (int i=0; i<nums.length; i++) { if (nums[i] == 0) { continue; } int j = i, k = getIndex(i, nums); // 检查同方向 while (nums[i] * nums[k] > 0 && nums[i] * nums[getIndex(k, nums)] > 0) { if (j == k) { // 碰到了 // 检查是否只有一个元素 if (j == getIndex(j, nums)) { break; } return true; } // 一前一后两个指针 j = getIndex(j, nums); k = getIndex(getIndex(k, nums), nums); } // 把已经走过的,置为0 j = i; k = nums[j]; // 但要注意不同方向的不要置为0 while (nums[j]*k > 0) { int tmp = getIndex(j, nums); nums[j] = 0; j = tmp; } } return false; } // 下面这个函数也是参考了上面Discuss里面的解法 private int getIndex(int i, int[] nums) { int n = nums.length; int j = (i + nums[i]) >= 0 ? (i + nums[i]) % n : n + ((i + nums[i]) % n); return j; } } public class Main { public static void main(String[] args) throws InterruptedException { System.out.println("Hello!"); Solution solution = new Solution(); // Your Codec object will be instantiated and called as such: int[] nums = {2, -1, 1, -2, -2}; boolean ret = solution.circularArrayLoop(nums); System.out.printf("ret:%b ", ret); System.out.println(); } }