Add Strings

来源：互联网发布：c语言结构体链表编辑：程序博客网时间：2024/05/16 10:45

题目地址：https://leetcode.com/problems/add-strings/

Given two non-negative numbers num1 and num2 represented as string, return the sum of num1 and num2.

Note:

The length of both num1 and num2 is < 5100.
Both num1 and num2 contains only digits 0-9.
Both num1 and num2 does not contain any leading zero.
You must not use any built-in BigInteger library or convert the inputs to integer directly.

题目要将两个数字字符串转换为数字，然后相加，得出结果，再转换为字符串。题目第四条说的很明确，不能用内置的转换方法，要是可以，那不就太简单了么。

题目本身没啥难度，关键是进位处理，这一点都考虑到了，也就没啥问题了。

public class AddStrings {    public String addStrings(String num1, String num2) {        StringBuilder result = new StringBuilder("");        //进位标记        boolean carry = false;        for (int i = num1.length() - 1, j = num2.length() - 1 ; i >= 0 && j >= 0; i--, j--) {            int t = 0;            if (carry)                t = num1.charAt(i) - '0' + num2.charAt(j) - '0' + 1;            else                t = num1.charAt(i) - '0' + num2.charAt(j) - '0';            if (t >= 10)                carry = true;            else                carry = false;            if (carry)                result.insert(0, String.valueOf((char)('0' + (t-10))));            else                result.insert(0, String.valueOf((char)('0' + t)));        }        // 不等长的时候，余下的位的计算，注意这里也有可能产生进位。        if (num1.length() > num2.length()) {            for (int i = num1.length() - num2.length() - 1; i >= 0; i--) {                int t = 0;                if (carry)                    t = num1.charAt(i) - '0' + 1;                else                    t = num1.charAt(i) - '0';                if (t >= 10)                    carry = true;                else                    carry = false;                if (carry)                    result.insert(0, String.valueOf((char)('0' + (t-10))));                else                    result.insert(0, String.valueOf((char)('0' + t)));            }        }        if (num2.length() > num1.length()) {            for (int i = num2.length() - num1.length() - 1; i >= 0; i--) {                int t = 0;                if (carry)                    t = num2.charAt(i) - '0' + 1;                else                    t = num2.charAt(i) - '0';                if (t >= 10)                    carry = true;                else                    carry = false;                if (carry)                    result.insert(0, String.valueOf((char)('0' + (t-10))));                else                    result.insert(0, String.valueOf((char)('0' + t)));            }        }        if (carry)            result.insert(0, '1');        return result.toString();    }    public static void main(String[] args) {        AddStrings addStrings = new AddStrings();        System.out.println(addStrings.addStrings("999929", "72"));    }}

时间复杂度为：O(max{m, n})，其中m和n分别是两个字符串的长度。

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