[leetcode 43][字符串相乘]

字符串相乘

• 给定两个以字符串形式表示的非负整数 num1 和 num2，返回 num1 和 num2 的乘积，它们的乘积也表示为字符串形式。

示例 1:
输入: num1 = “2”, num2 = “3”
输出: “6”
示例 2:
输入: num1 = “123”, num2 = “456”
输出: “56088”

class Solution {
    public String multiply(String num1, String num2) {
        if (num1.equals("0") || num2.equals("0")) {
            return "0";
        }
        List<LinkedList<String>> list = new ArrayList<>();
        int index = 0;
        for (int i = num1.length() - 1; i >= 0; i--) {
            LinkedList<String> subList = new LinkedList<>();
            int carry = 0;
            for (int j = num2.length() - 1; j >= 0; j--) {
                if (j == num2.length() - 1) {
                    for (int k = 0; k < index; k++) {
                        subList.add("#");
                    }
                }
                int num = Integer.parseInt(String.valueOf(num1.charAt(i))) * Integer.parseInt(String.valueOf(num2.charAt(j))) + carry;
                subList.addFirst(String.valueOf(num % 10));
                carry = num / 10;
            }
            if (carry != 0) {
                subList.addFirst(String.valueOf(carry));
            }
            list.add(subList);
            index++;
        }
        String result = "";
        int carry = 0;
        while (true) {
            int sum = 0;
            boolean flag = false;
            for (LinkedList<String> subList : list) {
                if (subList.size() != 0) {
                    String n = subList.removeLast();
                    if (!n.equals("#")) {
                        sum += Integer.parseInt(n);
                    }
                    flag = true;
                }
            }
            if (!flag) {
                break;
            }
            sum += carry;
            carry = sum / 10;
            result = String.valueOf(sum % 10) + result;
        }
        if (carry != 0) {
            result = String.valueOf(carry) + result;
        }
        return result;
    }
}

标签：String,num2,int,43,subList,相乘,carry,leetcode,num1
来源： https://blog.csdn.net/pp1981002445/article/details/120460351