【LeetCode】415 Add Strings （java实现）

原题链接

https://leetcode.com/problems/add-strings/

原题

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.

题目要求

题目叫“字符串相加”，给定两个非负的由字符串表示整数，求出它们的和。这里需要注意的是，num1和num2字符串的长度都小于5100，这是个很大的数字，肯定不能用int来计算。数字只包含0-9，也就是说没有负号，没有小数点等，完全当做整形来计算。不允许使用内建的大整形的库函数。

解法

实现一：思路比较简单，既然不能把字符串转为整形来计算，那就一个字符一个字符取出来相加，算好进位即可。

public String addStrings(String num1, String num2) {
    String bigStr = null;
    String smallStr = null;
    
    if (num1.length() >= num2.length()) {
        bigStr = num1;
        smallStr = num2;
    } else {
        bigStr = num2;
        smallStr = num1;
    }
    
    int big = bigStr.length();
    int small = smallStr.length();
    int carry = 0;
    char []ra = new char[big + 1];
    for (int i = 0; i < small; i++) {
        int b = Character.getNumericValue(bigStr.charAt(big - i - 1));
        int s = Character.getNumericValue(smallStr.charAt(small - i - 1));
        ra[big - i] = Character.forDigit((b + s + carry) % 10, 10);
        carry = (b + s + carry) / 10; 
    }

    for (int i = 0; i < big - small; i++) {
        int b = Character.getNumericValue(bigStr.charAt(big - small - i - 1));
        ra[big - small - i] = Character.forDigit((b+ carry) % 10, 10);
        carry = (b + carry)/ 10; 
    }
    if (carry != 0) {
        ra[0] = Character.forDigit(carry % 10, 10);
    }
    
    String ret = new String(ra).trim();
    return ret;
}

实现二：从disguss选出来的最简解，我把它从c++翻译成了Java。与实现一并没有什么本质不同，只是代码更简洁，更清晰。牛人太多，佩服。

public String addStrings(String num1, String num2) {
    int i = num1.length() - 1, j = num2.length() - 1, carry = 0;
    String res = "";
    while (i >= 0 || j >= 0) {
        if (i >= 0)
            carry += num1.charAt(i--) - '0';
        if (j >= 0)
            carry += num2.charAt(j--) - '0';
        res = Integer.toString(carry % 10) + res;
        carry /= 10;
    }
    return carry != 0 ? "1" + res : res;
}

来源：oschina

链接：https://my.oschina.net/u/1048295/blog/775837