Leetcode: Circular Array Loop

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?

注意The loop must be "forward" or "backward'. 所以这就是为什么[-2, 1, -1, -2, -2]是false的原因

Just think it as finding a loop in Linkedlist, except that loops with only 1 element do not count. Use a slow and fast pointer, slow pointer moves 1 step a time while fast pointer moves 2 steps a time. If there is a loop (fast == slow), we return true, just except there's only 1 element in the loop. else if we meet element with different directions, then the search fail, we set all elements along the way to 0. Because 0 is fail for sure so when later search meet 0 we know the search will fail.

public class Solution {
    int[] arr;
    public boolean circularArrayLoop(int[] nums) {
        arr = nums;
        for (int i=0; i<nums.length; i++) {
            if (nums[i] == 0) continue;
            //two pointers
            int j=i, k=shift(i);
            while (nums[i]*nums[k]>0 && nums[i]*nums[shift(k)]>0) {
                if (j == k) {
                    // check for loop with only one element
                    if (j == shift(j)) break;
                    return true;
                }
                j = shift(j);
                k = shift(shift(k));
            }
            // loop not found, set all element along the way to 0
            j = i;
            int val = nums[i];
            while (nums[j]*val > 0) {
                int next = shift(j);
                nums[j] = 0;
                j = next;
            }
        }
        return false;
    }
    
    public int shift(int pos) {
        return (pos+arr[pos]+arr.length) % arr.length;
    }
}

 Solution 2: 更清晰，没有找到Loop就set为0不是必需的

public class Solution {
    int[] arr;
    public boolean circularArrayLoop(int[] nums) {
        arr = nums;
        for (int i=0; i<nums.length; i++) {
            if (nums[i] == 0) continue;
            //two pointers
            int j=i, k=i;
            while (nums[i]*nums[shift(k)]>0 && nums[i]*nums[shift(shift(k))]>0) {
                j = shift(j);
                k = shift(shift(k));
                if (j == k) {
                    // check for loop with only one element
                    if (j == shift(j)) break;
                    return true;
                }
            }
            // loop not found, set all element along the way to 0， not necessary

        }
        return false;
    }
    
    public int shift(int pos) {
        return (pos+arr[pos]+arr.length) % arr.length;
    }
}