Question

457. Circular Array Loop

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Description

You are playing a game involving a circular array of non-zero integers nums . Each nums[i] denotes the number of indices forward/backward you must move if you are located at index i :

• If nums[i] is positive, move nums[i] steps forward, and
• If nums[i] is negative, move nums[i] steps backward.

Since the array is circular, you may assume that moving forward from the last element puts you on the first element, and moving backwards from the first element puts you on the last element.

A cycle in the array consists of a sequence of indices seq of length k where:

• Following the movement rules above results in the repeating index sequence seq[0] -> seq[1] -> ... -> seq[k - 1] -> seq[0] -> ...
• Every nums[seq[j]] is either all positive or all negative.
• k > 1

Return true _ if there is a cycle in _ nums _, or _ false _ otherwise_.

Example 1:

Input: nums = [2,-1,1,2,2]
Output: true
Explanation: The graph shows how the indices are connected. White nodes are jumping forward, while red is jumping backward.
We can see the cycle 0 --> 2 --> 3 --> 0 --> ..., and all of its nodes are white (jumping in the same direction).

Example 2:

Input: nums = [-1,-2,-3,-4,-5,6]
Output: false
Explanation: The graph shows how the indices are connected. White nodes are jumping forward, while red is jumping backward.
The only cycle is of size 1, so we return false.

 

Constraints:

• 1 <= nums.length <= 5000
• -1000 <= nums[i] <= 1000
• nums[i] != 0

Follow up: Could you solve it in O(n) time complexity and O(1) extra space complexity?

Solutions

Solution 1

class Solution:
  def circularArrayLoop(self, nums: List[int]) -> bool:
    n = len(nums)

    def next(i):
      return (i + nums[i] % n + n) % n

    for i in range(n):
      if nums[i] == 0:
        continue
      slow, fast = i, next(i)
      while nums[slow] * nums[fast] > 0 and nums[slow] * nums[next(fast)] > 0:
        if slow == fast:
          if slow != next(slow):
            return True
          break
        slow, fast = next(slow), next(next(fast))
      j = i
      while nums[j] * nums[next(j)] > 0:
        nums[j] = 0
        j = next(j)
    return False

class Solution {
  private int n;
  private int[] nums;

  public boolean circularArrayLoop(int[] nums) {
    n = nums.length;
    this.nums = nums;
    for (int i = 0; i < n; ++i) {
      if (nums[i] == 0) {
        continue;
      }
      int slow = i, fast = next(i);
      while (nums[slow] * nums[fast] > 0 && nums[slow] * nums[next(fast)] > 0) {
        if (slow == fast) {
          if (slow != next(slow)) {
            return true;
          }
          break;
        }
        slow = next(slow);
        fast = next(next(fast));
      }
      int j = i;
      while (nums[j] * nums[next(j)] > 0) {
        nums[j] = 0;
        j = next(j);
      }
    }
    return false;
  }

  private int next(int i) {
    return (i + nums[i] % n + n) % n;
  }
}

class Solution {
public:
  bool circularArrayLoop(vector<int>& nums) {
    int n = nums.size();
    for (int i = 0; i < n; ++i) {
      if (!nums[i]) continue;
      int slow = i, fast = next(nums, i);
      while (nums[slow] * nums[fast] > 0 && nums[slow] * nums[next(nums, fast)] > 0) {
        if (slow == fast) {
          if (slow != next(nums, slow)) return true;
          break;
        }
        slow = next(nums, slow);
        fast = next(nums, next(nums, fast));
      }
      int j = i;
      while (nums[j] * nums[next(nums, j)] > 0) {
        nums[j] = 0;
        j = next(nums, j);
      }
    }
    return false;
  }

  int next(vector<int>& nums, int i) {
    int n = nums.size();
    return (i + nums[i] % n + n) % n;
  }
};

func circularArrayLoop(nums []int) bool {
  for i, num := range nums {
    if num == 0 {
      continue
    }
    slow, fast := i, next(nums, i)
    for nums[slow]*nums[fast] > 0 && nums[slow]*nums[next(nums, fast)] > 0 {
      if slow == fast {
        if slow != next(nums, slow) {
          return true
        }
        break
      }
      slow, fast = next(nums, slow), next(nums, next(nums, fast))
    }
    j := i
    for nums[j]*nums[next(nums, j)] > 0 {
      nums[j] = 0
      j = next(nums, j)
    }
  }
  return false
}

func next(nums []int, i int) int {
  n := len(nums)
  return (i + nums[i]%n + n) % n
}