根据周X HH:mm:ss判断时间是否交叉重叠

根据如下itemInfo这种数据判断时间是否冲突,我的思路是将时间全部转化为秒,判别区间是否交叉重叠,是即冲突。
代码如下(待优化)

itemInfo:[
  { 
    period: ['03:20:10','15:20:10'],
    weekList: ['周一','周三','周四']
  },
  { 
    period: ['02:20:15','03:30:30'],
    weekList: ['周三','周五']
  },
  { 
    period: ['02:20:15','03:30:30'],
    weekList: ['周二']
  },
  { 
    period: ['16:20:15','20:30:30'],
    weekList: ['周一','周五']
  }
],
// 周转化数字
    getWeekTransform(week){ 
      let newweek = week.map(item=>{ 
        if(item=='周一'){ 
          return item=1
        }else if(item=='周二') { 
          return item=2
        }else if(item=='周三') { 
          return item=3
        }else if(item=='周四') { 
          return item=4
        }else if(item=='周五') { 
          return item=5
        }
      })
      return newweek
    },
    getHourTime(val){  // 将'12:30:30'格式的时间换算为秒
      let hour = val.substring(0,2)
      let minute = val.substring(3,5)
      let second = val.substring(6,8)
      let sum = Number(hour)*3600+Number(minute)*60+Number(second)
      return sum
    },
    getSelfInterval(a,b){  //a为['16:20:15','20:30:30']时间区间,b为['周一','周三','周四']格式换算后的[1,3,4]数组
      let selfInterval = b.map(item=>{   //时间段换算成区间
        let sum = []
        sum.push(Number(item)*86400 + this.getHourTime(a[0]))
        sum.push(Number(item)*86400 + this.getHourTime(a[1]))
        return sum
      })
      console.log(selfInterval)
      return selfInterval
    },
    // 判断数字区间是否存在重叠交叉
    getNumInterval(a,b){ 
      let max=[]; // 存储两个数字区间start,如[1000,2000]、[3000,4000]存[1000,3000]
      let min=[]; // 存储两个数字区间end,如[1000,2000]、[3000,4000]存[2000,4000]
      max.push(a[0])
      max.push(b[0])
      min.push(a[1])
      min.push(b[1])
      if(Math.max.apply(null, max) <= Math.min.apply(null, min)){ 
        console.log("区间存在重叠交叉!");
      }
    },
    getItemInfo(){ 
      let arr = this.itemInfo.map(item=>{ 
        return this.getSelfInterval(item.period,this.getWeekTransform(item.weekList))
      })
      let allInterval = arr.flat(1) //所有区间数组
      for(let i = 0;i < allInterval.length; i++){ 
        for(let j = 0;j < allInterval.length; j++){ 
          if(i!==j){   //for循环对比区间,去除自身对比
            this.getNumInterval(allInterval[i],allInterval[j])
          }
        }
      }
    },

写的有些烂,各位大佬多提意见。

    原文作者:汪酱~
    原文地址: https://blog.csdn.net/qq_44768564/article/details/118559786
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