Google/LintCode：E-Big Integer Addition

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题目

题目来源：Link

Given two non-negative integers num1 and num2represented as string, return the sum of num1 and num2.

注意事项

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.

您在真实的面试中是否遇到过这个题？

Yes

样例

Given num1 = "123", num2 = "45"
return "168"

代码

（1）用Java内置BigInteger

import java.math.BigInteger;public class Solution {    /**     * @param num1 a non-negative integers     * @param num2 a non-negative integers     * @return return sum of num1 and num2     */    public String addStrings(String num1, String num2) {        // Write your code here        BigInteger bnum1 = new BigInteger(num1);        BigInteger bnum2 = new BigInteger(num2);        BigInteger sum = bnum1.add(bnum2);        return sum.toString();    }}

（2）自己实现大数

import java.math.BigInteger;public class Solution {    /**     * @param num1 a non-negative integers     * @param num2 a non-negative integers     * @return return sum of num1 and num2     */    public String addStrings(String num1, String num2) {        // Write your code here        int n1 = num1.length();        int n2 = num2.length();        int n = Math.min(n1,n2);        String rnum1 = new StringBuffer(num1).reverse().toString();        String rnum2 = new StringBuffer(num2).reverse().toString();        String res="";        int up = 0;        int i=0;        for(; i<n; i++){            int sum = Integer.parseInt(String.valueOf(rnum1.charAt(i)))+Integer.parseInt(String.valueOf(rnum2.charAt(i)))+up;            up = sum/10;            int re = sum%10;            res = res + re;        }        while(i<n1){            int sum = Integer.parseInt(String.valueOf(rnum1.charAt(i)))+up;            up = sum/10;            int re = sum%10;            res = res + re;            i++;        }        while(i<n2){            int sum = Integer.parseInt(String.valueOf(rnum2.charAt(i)))+up;            up = sum/10;            int re = sum%10;            res = res + re;            i++;        }        //忘记了这个case        //最后可能还有进位        if(up!=0)            res+=up;                    return new StringBuffer(res).reverse().toString();    }}

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