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=; i<nums.length; i++) {

            if (nums[i] == ) {

                continue;

            }

            int j = i, k = getIndex(i, nums);

            // 检查同方向

            while (nums[i] * nums[k] >  && nums[i] * nums[getIndex(k, nums)] > ) {

                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 > ) {

                int tmp = getIndex(j, nums);

                nums[j] = ;

                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]) >=  ? (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 = {, -, , -, -};

        boolean ret = solution.circularArrayLoop(nums);

        System.out.printf("ret:%b\n", ret);

        System.out.println();

    }

}