判断如下时间是否有交集

const time = [
      {
        startTime: "17:31",
        endTime: "17:35"
      },
      {
        startTime: "17:35",
        endTime: "17:39"
      },
      {
        startTime: "18:00",
        endTime: "18:39"
      },
      {
        startTime: "19:35",
        endTime: "19:39"
      }
    ]

  const date = [
    {
      startDate: "2019-08-14",
      endDate: "2019-08-17"
    },
    {
      startDate: "2019-08-17",
      endDate: "2019-08-20"
    }
  ]

判断交集的方法

 const check = (a,b,x,y) => {
      if(y<a||b<x){  
        return false;
      }else{
        return true;
      }
    } 
  
    const checkTime = (a,b,x,y) => {
      let times1=[];
      let times2 = [];
      if (a<b) {
        //未跨天
        times1.push([a,b]);
      }else{
        //跨天
        times1.push([a,"24:00"],["00:00",b]);
      }
      
      if (x<y) {
        times2.push([x,y]);
      }else{
        times2.push([x,"24:00"],["00:00",y]);
      }
      
      let flag = false;
      //循环比较时间段是否冲突
      for (let i = 0; i < times1.length; i++) {
        if (flag) break;
        for (let j = 0; j < times2.length; j++) {
          if(check(times1[i][0],times1[i][1],times2[j][0],times2[j][1])){
            flag = true;
            break;
          }
        }
      }
      if (flag) {
        return {
          flag,
          msg:`发生冲突: ${times1[0][0]}-${times1[0][1]} 与 ${times2[0][0]}-${times2[0][1]}`
        };
      }else{
        return {
          flag,
          msg:`发生冲突: ${times1[0][0]}-${times1[0][1]} 与 ${times2[0][0]}-${times2[0][1]}`
        };
      }
    }

循环判断每个对象是否有交集

// 判断时间是否有交集
        for (let i = 0;i < time.length;i++){
          const currentA = time[i];
          for (let j = 0;j < time.length;j++){
            if(i === j){
              continue;
            }
            const currentB = time[j]
            const res = checkTime(currentA.startTime,currentA.endTime,currentB.startTime,currentB.endTime);
            if(res.flag){
              // 弹出报错
              console.log(res.msg);
              return;
            }
          }
        }