一道面试题论数据结构的重要性

导读

机试题，两个字符串相加，好的解法我在沸点中看到过，还研究了下，结果忘了。。。

https://codepen.io/xiaodun/pen/dBWByw

image

笨重的实现

机试的时候比下面的更长，因为要兼容IE8！

    function strNumAddAble(argNum1 = "", argNum2 = "") {
      if (argNum1.length < argNum2.length) {
        [argNum1, argNum2] = [argNum2, argNum1];
      }

      //在前面补0 保证两个字符串长度一致
      argNum2 = argNum2.padStart(argNum1.length, "0");

      //进位
      let addNum = 0;

      //存储每次计算的结果
      let list = [];
      for (let i = argNum1.length - 1; i >= 0; i--) {
        let one = argNum1[i];
        let two = argNum2[i];
        let result = +one + +two + addNum;

        addNum = (result / 10) | 0;
        let item = result % 10;
        list.unshift(item);
      }
      //循环本省不能处理99 + 1的情况
      if (addNum) {
        list.unshift("1");
      }
      return list.join("");
    }

    console.log(strNumAddAble("10", "2"));
    console.log(strNumAddAble("1128", "2"));
    console.log(strNumAddAble("939", "1"));
  

上述的很多逻辑用在处理argNum1,argNum2的长度上，进位的原因，还要再for循环外面处理一下。

好的实现

展现了数组的强大以及while循环的优势

    function strNumAddAble(argNum1 = "", argNum2 = "") {
      let num1List = [...argNum1],
        num2List = [...argNum2];

      let isCarry = false;
      let result = "";
      while (num1List.length || num2List.length || isCarry) {
        let item = ~~num2List.pop() + ~~num1List.pop() + isCarry;

        result = (item % 10) + result;
        isCarry = item > 9;
      }

      return result;
    }

    console.log(strNumAddAble("0", "2"));
    console.log(strNumAddAble("1128", "2"));
    console.log(strNumAddAble("999", "1"));

~~undefined 全等于0

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