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  • 数组合并

     一:合并后的数组为:['A1','A2','A',B1','B2','B','C1','C2','C','D1','D2','D']
        let ary1 = ['A1', 'A2', 'B1', 'B2', 'C1', 'C2', 'D1', 'D2'];
        let ary2 = ['A', 'B', 'C', 'D'];
     
        ary2 = ary2.map(item => item + '3');
        let arr = ary1.concat(ary2);
        arr = arr.sort((a, b) => a.localeCompare(b)).map(item => {
          return item.replace('3', '');
        })
    二:合并后的数组为:['D1','D2','D','A1','A2','A','C1','C2','C','B1','B2','B']
        let ary1 = ['D1', 'D2', 'A1', 'A2', 'C1', 'C2', 'B1', 'B2'];
        let ary2 = ['B', 'A', 'D', 'C'];

        let n = 0;
        for (let i = 0; i < ary2.length; i++) {
          let item2 = ary2[i];
          for (let k = 0; k < ary1.length; k++) {
            let item1 = ary1[k];
            if (item1.includes(item2)) {
              // => 如果包含就记录一下当前这一项的索引位置(后面还有包含的会重新记录这个值)
              n = k;
            }
          }
          // => 把当前 item2 这一项插入到 ary1 中 n 的后面
           ary1.splice(n + 1, 0, item2);
         }

        console.log(ary1)
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  • 原文地址:https://www.cnblogs.com/HYTing/p/12621031.html
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