LeetCode 43

https://leetcode-cn.com/problems/multiply-strings/

这个题没啥好说的，我是直接暴力法，开了个n*m的数组，n为两个数字中最长的那个的长度，m是两个数字长度之和去存放计算过程中的中间数，然后再进行加法操作即可。时间效率比较低，空间效率也很低，但是AC了。。。

class Solution {
    public String multiply(String num1, String num2) {
        if(num1 == null || num2 == null || num1.length() == 0 || num2.length() == 0){
            return "";
        }
        if("0".equals(num1) || "0".equals(num2)){
            return "0";
        }
        char[] str1 = num1.toCharArray();
        char[] str2 = num2.toCharArray();
        int totalLength = str1.length + str2.length;
        int bigger = Math.max(str1.length, str2.length);
        int[][] temp = new int[bigger][totalLength];
        int count = 0;
        int layer = 0;
        int i = str1.length - 1;
        int j = str2.length - 1;
        for (; i >= 0; i--) {
            int k = 0;
            for (j = str2.length - 1; j >= 0; j--) {
                int realNum = (str2[j] - '0') * (str1[i] - '0');
                temp[bigger - i - 1][totalLength - layer - 1 - k] = realNum % 10 + count;
                count = realNum / 10;
                k++;
            }
            temp[bigger - i - 1][totalLength - layer - 1 - k] = count;
            count = 0;
            layer++;
        }
        StringBuilder sb = new StringBuilder();
        int sum = 0;
        for (j = totalLength - 1; j >= 0; j--) {
            for (i = 0; i < bigger; i++) {
                sum += temp[i][j];
            }
            sum += count;
            count = sum / 10;
            while (sum >= 10) {
                sum %= 10;
            }
            sb.append(sum);
            sum = 0;
        }
        if (count != 0) {
            sb.append(count);
        }
        return sb.reverse().toString().replaceFirst("^0*","");
    }
}

评论区中有更快的方法，直接开一维数组存中间过程的数即可，根本用不着吧所有中间过程的数保存下来后再做加法运算。。。。。。。。