Question

1550. Three Consecutive Odds

中文文档

Description

Given an integer array arr, return true if there are three consecutive odd numbers in the array. Otherwise, return false.

 

Example 1:

Input: arr = [2,6,4,1]
Output: false
Explanation: There are no three consecutive odds.

Example 2:

Input: arr = [1,2,34,3,4,5,7,23,12]
Output: true
Explanation: [5,7,23] are three consecutive odds.

 

Constraints:

• 1 <= arr.length <= 1000
• 1 <= arr[i] <= 1000

Solutions

Solution 1

class Solution:
  def threeConsecutiveOdds(self, arr: List[int]) -> bool:
    cnt = 0
    for v in arr:
      if v & 1:
        cnt += 1
      else:
        cnt = 0
      if cnt == 3:
        return True
    return False

class Solution {
  public boolean threeConsecutiveOdds(int[] arr) {
    int cnt = 0;
    for (int v : arr) {
      if (v % 2 == 1) {
        ++cnt;
      } else {
        cnt = 0;
      }
      if (cnt == 3) {
        return true;
      }
    }
    return false;
  }
}

class Solution {
public:
  bool threeConsecutiveOdds(vector<int>& arr) {
    int cnt = 0;
    for (int v : arr) {
      if (v & 1)
        ++cnt;
      else
        cnt = 0;
      if (cnt == 3) return true;
    }
    return false;
  }
};

func threeConsecutiveOdds(arr []int) bool {
  cnt := 0
  for _, v := range arr {
    if v%2 == 1 {
      cnt++
    } else {
      cnt = 0
    }
    if cnt == 3 {
      return true
    }
  }
  return false
}

function threeConsecutiveOdds(arr: number[]): boolean {
  let cnt = 0;
  for (const v of arr) {
    if (v & 1) {
      ++cnt;
    } else {
      cnt = 0;
    }
    if (cnt == 3) {
      return true;
    }
  }
  return false;
}

Solution 2

class Solution:
  def threeConsecutiveOdds(self, arr: List[int]) -> bool:
    for i in range(len(arr) - 2):
      if arr[i] % 2 + arr[i + 1] % 2 + arr[i + 2] % 2 == 3:
        return True
    return False