如何判断数组中start_time与end_time相同组合的出现次数是否达到3次

• 以start_time和end_time拼接的字符串作为唯一标识，用Map/普通对象统计每个时间组合的出现次数
• 遍历数组的过程中实时更新计数，只要某组计数达到3就直接返回false，提前终止遍历提升性能
• 全量遍历完成后无符合条件的组合则返回true

function validateTimeArr(arr) {
  const countMap = new Map()
  for (const item of arr) {
    // 用双竖线作为分隔符避免特殊字符导致的key冲突
    const key = `${item.start_time}||${item.end_time}`
    const currentCount = (countMap.get(key) || 0) + 1
    if (currentCount >= 3) {
      return false
    }
    countMap.set(key, currentCount)
  }
  return true
}

用你给出的示例数组调用函数即可验证效果：

const testArr = [
  {"id": 1,"start_time": "10:00","end_time": "11:00"},
  {"id": 2,"start_time": "11:00","end_time": "12:00"},
  {"id": 3,"start_time": "13:00","end_time": "15:00"},
  {"id": 4,"start_time": "13:00","end_time": "14:00"},
  {"id": 5,"start_time": "13:00","end_time": "14:00"},
  {"id": 6,"start_time": "13:00","end_time": "14:00"},
  {"id": 7,"start_time": "14:00","end_time": "15:00"},
  {"id": 8,"start_time": "17:00","end_time": "18:00"},
  {"id": 9,"start_time": "17:00","end_time": "18:00"},
  {"id": 10,"start_time": "17:00","end_time": "18:00"}
]

console.log(validateTimeArr(testArr)) // 输出 false

如果是其他编程语言，思路完全一致，替换对应哈希表的实现即可，无需调整核心逻辑。

内容的提问来源于stack exchange，提问作者Wcan