LeetCode 43. Multiply Strings

自己来实现字符串的 mul函数, 给定的字符串符合数字的要求,very simple Solution
考虑清楚中间的carry, index 就行了.

class Solution {
    public String multiply(String num1, String num2) {
        if (num1.equals("0") || num2.equals("0")) {
            return "0";
        }
        char[] char1 = num1.toCharArray();
        char[] char2 = num2.toCharArray();
        int len1 = num1.length();
        int len2 = num2.length();
        int[] res = new int[len1 + len2];
        for (int i = len1 - 1; i >= 0; i--) {
            int carry = 0; // 进位
            int index = res.length - 1 - (len1 - 1 - i); // 相对坐标
            for (int j = len2 - 1; j >= 0; j--) {
                int mul = res[index] + (char1[i] - '0')*(char2[j] - '0') + carry;
                carry = mul/10;
                res[index] = mul%10;
                index--;
            }
            res[index] += carry; // 如果有进位的话
        }
        StringBuilder sb = new StringBuilder();
        boolean flag = true;
        for (int i = 0; i < res.length; i++ ) {
            if (res[i] == 0 && flag) {
                continue;
            }
            flag = false;
            sb.append(res[i] + "");
        }
        return sb.toString();
    }
}

Python is just a joke :blush:

class Solution:
    def multiply(self, num1, num2):
        """
        :type num1: str
        :type num2: str
        :rtype: str
        """
        return str(int(num1)*int(num2))

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转载自blog.csdn.net/qq_33797928/article/details/80096420