算法问题-大数相加

大数相加

比如远超long的数字相加，求其结果，需要用到数组，通过进位去实现

package algorithm.bignumbersum;

public class BigNumberSum {
    /**
     * 字符数组翻转
     * 12345 3   04  13 22
     * 1234 2   03  12
     *
     * 同时进行末尾补零，方便下面的运算
     */
    public static int[] reverse(String str, int maxLen){
        char[] chars = str.toCharArray();
        int[] array = new int[maxLen];
        for (int i = chars.length-1  ; i >=0; i--) {
            array[chars.length-i-1] = chars[i]-'0';
        }

        return array;
    }

    //去掉翻转后高位的零
    public static String reverse(int[] ints){
        StringBuffer sb = new StringBuffer();
        for (int i = ints.length-1; i >=0; i--) {
            if(ints[i]!=0){
                sb.append(ints[i]);
            }else{
                continue;
            }
        }
        return sb.toString();
    }


    /**
     * 大整数相加
     * 思路拆分成数组 然后进位
     */
    public static String bigNumberSum(String num1,String num2){
        /**
         * 先把数组翻转
         * 12345+123 可以转换成 54321+321 从左向右做加法
         *
         */
        int maxLen = num1.length()>num2.length()?num1.length():num2.length();
        int[] n1 = reverse(num1,maxLen);
        int[] n2 = reverse(num2,maxLen);
        int[] res = new int[maxLen+1];

        for (int i = 0; i < maxLen; i++) {
            int temp = n1[i] + n2[i];
            //考虑 十进制 进位问题
            if(temp>=10) {
                res[i + 1] = 1;
                temp -=10;
            }
            res[i]=temp;
        }
        return reverse(res);
    }

    public static void main(String[] args) {
        //12345+123 = 12468
        System.out.println(bigNumberSum("12345","123"));
    }
}