leetcode oj java Add Strings

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一、问题描述：

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.

二、解决思路：

不能转化为integer，就用char数组，因为char之间直接相加返回的是ascii码。两个char相加再减去48*2 就得到十进制的结果。（0-9的ascii是48-57）

剩下的就是实现基本的十进制加法啦。要考虑的因素比较多~ 一定要细心

三、代码：

public class Solution {    public String addStrings(String num1, String num2) {        String res = "";        char[] c1 = num1.toCharArray();        char[] c2 = num2.toCharArray();        int temp = 0;        int re = 0;        for (int i = c1.length - 1, j = c2.length - 1; i > -1 && j > -1; i--, j--) {            char cn1 = c1[i];            char cn2 = c2[j];            re = cn1 + cn2 + temp - 48 * 2;            if (re > 9) {                temp = re / 10;                re = re % 10;            } else {                temp = 0;            }            res = re + res;;        }        if (c1.length == c2.length && temp > 0) {            res = temp + res;            return res;        }        if (c1.length - c2.length > 1) {            re = c1[c1.length - c2.length - 1] + temp - 48;            if (re > 9) {                temp = re / 10;                re = re % 10;            } else {                temp = 0;            }            res = re + res;            for (int i = c1.length - c2.length - 2; i > -1; i--) {                re = c1[i] + temp - 48;                if (re > 9) {                    temp = re / 10;                    re = re % 10;                } else {                    temp = 0;                }                res = re + res;            }            if (temp > 0) {                res = temp + res;            }        }        if (c1.length - c2.length == 1) {            re = c1[c1.length - c2.length - 1] + temp - 48;            res = re + res;        }        if (c2.length - c1.length > 1) {            re = c2[c2.length - c1.length - 1] + temp - 48;            if (re > 9) {                temp = re / 10;                re = re % 10;            } else {                temp = 0;            }            res = re + res;            for (int i = c2.length - c1.length - 2; i > -1; i--) {                re = c2[i] + temp - 48;                if (re > 9) {                    temp = re / 10;                    re = re % 10;                } else {                    temp = 0;                }                res = re + res;            }            if (temp > 0) {                res = temp + res;            }        }        if (c2.length - c1.length == 1) {            re = c2[c2.length - c1.length - 1] + temp - 48;            res = re + res;        }        return res;    }}

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