兩個(gè)string,里面存的都是數(shù)字該如何相加？

1、使用BigInteger類、BigDecimal類

public static void main(String[] args) {
        String a="19238248931244327089";
        String b="43109431029897431";
        String str=new BigInteger(a).add(new BigInteger(b)).toString();
        System.out.println(str);
    }

2、反轉(zhuǎn)字符串、對(duì)齊字符串缺位補(bǔ)0、將兩個(gè)正整數(shù)相加

public static String add(String n1,String n2){

            String result="";

            //反轉(zhuǎn)字符串
            String num1=new StringBuffer(n1).reverse().toString();
            String num2=new StringBuffer(n2).reverse().toString();

            int len1=num1.length();
            int len2=num2.length();
            int maxLen=len1>len2?len1:len2;
            //定義和(可能)
            int nSum[]=new int[maxLen+1];

            boolean nOverFlow=false;

            //對(duì)齊字符串
            if(len1<len2){
                for (int i = len1; i < len2; i++) {
                    num1+="0";
                }
            }else if(len1>len2){
                for (int i = len2; i < len1; i++) {
                    num2+="0";
                }
            }

            //兩個(gè)數(shù)相加
            for (int i = 0; i < maxLen; i++) {
                //進(jìn)位數(shù)從第二次開始算
                if (nOverFlow) {
                    nSum[i]=Integer.parseInt(num1.charAt(i)+"")+
                            Integer.parseInt(num2.charAt(i)+"")+1;
                }else{
                    nSum[i]=Integer.parseInt(num1.charAt(i)+"")+
                            Integer.parseInt(num2.charAt(i)+"");
                }
                //處理溢出位
                nOverFlow=handleSumOverTen(nSum,i);

            }

            //處理最高位
            if(nOverFlow) {
                nSum[maxLen] = 1;
            }else {
                nSum[maxLen] =0 ;
            }

            for (int i = 0; i < nSum.length; i++) {
                result+=String.valueOf(nSum[i]);
            }
            String result1=new StringBuffer(result).reverse().toString();
            return result1;
        }

        private static boolean handleSumOverTen(int[] nSum, int i) {

            boolean flag = false;
            if(nSum[i] >= 10) {
                nSum[i] = nSum[i] - 10;
                flag = true;
            }
            else {
                flag = false;
            }
            return flag;
        }
       public static void main(String[] args) {
        String num=add("19238248931244327089", "43109431029897431");
        System.out.println(num);
        }

這個(gè)結(jié)果可能會(huì)多出0字符

3、補(bǔ)齊字符串(使用StringBuffere中的insert方法在字符串索引為0的位置插入len個(gè)0)、對(duì)齊相加

public class test {

    public static void main(String[] args) {
        int[] result = bigNumSum("19238248931244327089", "43109431029897431");
        for(int i=0; i < result.length; i++) {
            System.out.print(result[i]);
        }
    }

    public static int[] bigNumSum(String num1, String num2) {

        String number1 = num1;
        String number2 = num2;

        int len1=number1.length();
        int len2=number2.length();
        int len=Math.abs(len1-len2);
        char insertNum[]=new char[len];
        for (int i = 0; i < insertNum.length; i++) {
            insertNum[i]='0';
        }
        String str1="";
        String str2="";
        //補(bǔ)齊兩個(gè)字符串
        if (len1<len2) {

            str1=new StringBuffer(number1).insert(0, insertNum).toString();
            str2=number2;
        }else if(len1>len2){
            str1=number1;
            str2=new StringBuffer(number2).insert(0, insertNum).toString();
        }

        //字符串轉(zhuǎn)換成字符數(shù)組
        char[] ch1 = str1.toCharArray();
        char[] ch2 = str2.toCharArray();
        int[] sum;
        //為true時(shí)表示兩數(shù)相加>=10
        boolean flag = false;

        //相加結(jié)果的長度為任一長度+1，因?yàn)樽罡呶幌嗉涌赡?gt;10
        sum = new int[ch1.length+1];
        //從個(gè)位開始相加
        for(int i=ch1.length-1; i>=0; i--) {
            //如果上一次相加和大于1，本次相加結(jié)果加1
            if(flag) {
                //
                sum[i+1] = (int)(ch1[i] - '0') + (int)(ch2[i] - '0') + 1;
            }else {
                sum[i+1] = (int)(ch1[i] - '0') + (int)(ch2[i] - '0');
            }
            flag = handleSumOverTen(sum, i); //處理兩數(shù)相加是否>10
        }

        handleTopDigit(flag, sum); //處理最高位
        return sum;


    }

    /*
     * 處理兩數(shù)相加是否>10
     */
    public static boolean handleSumOverTen(int[] sum, int i) {
        boolean flag = false;
        if(sum[i+1] >= 10) {
            sum[i+1] = sum[i+1] - 10;
            flag = true;
        }
        else {
            flag = false;
        }
        return flag;
    }

    /*
     * 處理最高位
     */
    public static void handleTopDigit(Boolean flag, int[] sum) {
        if(flag) {
            sum[0] = 1;
        }else {
            sum[0] = 0;
        }
    }

}

4、此方法與三方法基本一致，不同之處是三中方法因?yàn)閷蓚€(gè)字符串長度通過補(bǔ)0相等而不需要分步進(jìn)行判斷，但整體運(yùn)行效率還是四分法高

public class test {

    public static void main(String[] args) {
        int[] result = bigNumSum("19238248931244327089", "43109431029897431");
        for(int i=0; i < result.length; i++) {
            System.out.print(result[i]);
        }
    }

    public static int[] bigNumSum(String num1, String num2) {

        String number1 = num1;
        String number2 = num2;
        //字符串轉(zhuǎn)換成字符數(shù)組
        char[] ch1 = number1.toCharArray();
        char[] ch2 = number2.toCharArray();
        int[] sum;
        //取位數(shù)之差
        int len = Math.abs(ch1.length - ch2.length);
        //為true時(shí)表示兩數(shù)相加>=10
        boolean flag = false;

        //如果兩個(gè)數(shù)的長度相等
        if(ch1.length == ch2.length) {

            //相加結(jié)果的長度為任一長度+1，因?yàn)樽罡呶幌嗉涌赡?gt;10
            sum = new int[ch1.length+1];
            //從個(gè)位開始相加
            for(int i=ch1.length-1; i>=0; i--) {
                //如果上一次相加和大于1，本次相加結(jié)果加1
                if(flag) {
                    //
                    sum[i+1] = (int)(ch1[i] - '0') + (int)(ch2[i] - '0') + 1;
                }else {
                    sum[i+1] = (int)(ch1[i] - '0') + (int)(ch2[i] - '0');
                }
                flag = handleSumOverTen(sum, i, len); //處理兩數(shù)相加是否>10
            }

            handleTopDigit(flag, sum); //處理最高位
            return sum;
        }
        else if(ch1.length > ch2.length) { //如果數(shù)1的長度大于數(shù)2的長度
            sum = new int[ch1.length+1]; //結(jié)果的長度為數(shù)1的長度+1

            for(int i=ch2.length-1; i>=0; i--) {
                if(flag) {
                    sum[i+len+1] = (int)(ch1[i+len] - '0') + (int)(ch2[i] - '0') + 1;
                }
                else {
                    sum[i+len+1] = (int)(ch1[i+len] - '0') + (int)(ch2[i] - '0');
                }

                flag = handleSumOverTen(sum, i, len);
            }

            for(int i=ch1.length-ch2.length-1; i>=0; i--) { //處理數(shù)1多出來的位數(shù)
                if(flag) {
                    sum[i+1] = (int)(ch1[i] - '0') + 1;
                }
                else {
                    sum[i+1] = (int)(ch1[i] - '0');
                }
                flag = handleSumOverTen(sum, i, 0);
            }

            handleTopDigit(flag, sum);
            return sum;
        }
        else {
            sum = new int[ch2.length+1];

            for(int i=ch1.length-1; i>=0; i--) {
                if(flag) {
                    sum[i+len+1] = (int)(ch1[i] - '0') + (int)(ch2[i+len] - '0') + 1;
                }
                else {
                    sum[i+len+1] = (int)(ch1[i] - '0') + (int)(ch2[i+len] - '0');
                }

                flag = handleSumOverTen(sum, i, len);
            }

            for(int i=ch2.length-ch1.length-1; i>=0; i--) {
                if(flag) {
                    sum[i+1] = (int)(ch2[i] - '0') + 1;
                }
                else {
                    sum[i+1] = (int)(ch2[i] - '0');
                }
                flag = handleSumOverTen(sum, i, 0);
            }

            handleTopDigit(flag, sum);
            return sum;
        }
    }

    /*
     * 處理兩數(shù)相加是否>10
     */
    public static boolean handleSumOverTen(int[] sum, int i, int len) {
        boolean flag = false;
        if(sum[i+len+1] >= 10) {
            sum[i+len+1] = sum[i+len+1] - 10;
            flag = true;
        }
        else {
            flag = false;
        }
        return flag;
    }

    /*
     * 處理最高位
     */
    public static void handleTopDigit(Boolean flag, int[] sum) {
        if(flag) {
            sum[0] = 1;
        }else {
            sum[0] = 0;
        }
    }

}