[Amazon] Big Integer Addition

最新推荐文章于 2021-08-27 10:10:00 发布

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最新推荐文章于 2021-08-27 10:10:00 发布

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分类专栏： mock interview 文章标签： leetcode 亚马逊面试 lintcode onsite

本文链接：https://blog.youkuaiyun.com/offerkkl/article/details/75045834

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本文介绍了一种处理两个非负整数字符串相加的方法，通过补零对齐位数，实现逐位相加并处理进位，最终返回两数之和的字符串形式。适用于整数长度不超过5100的情况。

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

Notice

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

Example

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

思路：把长度短的字符串用0补齐位数，从最后一位开始借位相加，反转即可。

分析："123"和"045"

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) {
        StringBuffer sb=new StringBuffer();
        int len=Math.max(num1.length(),num2.length());
        int carry=0;
        for(int i=len-1;i>=0;i--){
            if(num1.length()<num2.length()){
                for(int j=num1.length();j<num2.length();j++){
                    num1="0"+num1;
                }
            }
            if(num1.length()>num2.length()){
                for(int j=num2.length();j<num1.length();j++){
                    num2="0"+num2;
                }
            }
            char c1=num1.charAt(i);
            char c2=num2.charAt(i);
            int digit=(carry+(c1-'0')+(c2-'0'))%10;
            sb.append(digit);
            carry=(carry+(c1-'0')+(c2-'0'))/10;
        }
        
        if(carry!=0){
            sb.append(carry);
        }
        
        return sb.reverse().toString();
    }
}

第二遍

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) {
        int diff=Math.abs(num1.length()-num2.length());
        if(num1.length()>num2.length()){
            for(int i=0;i<diff;i++){
                num2="0"+num2;
            }
        }else{
            for(int i=0;i<diff;i++){
                num1="0"+num1;
            }
        }
        
        String result="";
        int carry=0;
        for(int i=num1.length()-1;i>=0;i--){
            int sum=carry+(num1.charAt(i)-'0')+(num2.charAt(i)-'0');          //要注意charAt之后要-'0'
            int digit=sum%10;
            result+=digit;
            carry=sum/10;
        }
        
        if(carry!=0){
            result+=carry;
        }
        return new StringBuffer(result).reverse().toString();
    }
}