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;
     }
 }