Java实现BigInteger大数相加

时常见到这个题，由于long型的变量有上限，不能进行大数操作，先转化为一个一个字符分别运算，加减乘除同理。

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/**
 * @author _yiyi
 * @create 2018/8/23
 */
public class BigNumInteger {
    public String number1 = "11524323423435";
    public String number2 = "922323";
    /***
     *
     * @param number1
     * @param number2
     * @return 结果字符串
     * 1,先找出字符串大小，然后使其两个字符串长度一致，短在前面补齐0
     * 2，按照加法运算原则从后面往前加，设置进位标识flag，进位加一。
     * 3，重复2运算到结束。
     */
    private String bigNumAdd(String number1, String number2) {
        int maxLength, minLength;
        int flag = 0;//进位标志
        StringBuilder temp = new StringBuilder();
        maxLength = number1.length() > number2.length() ? number1.length() : number2.length();
        minLength = number1.length() < number2.length() ? number1.length() : number2.length();
        String result = "";//结果集
        //先把2个不同长度的字符串补齐0，短的那一边前面补齐0,例如，1234,0012.
        for (int i = 0; i < maxLength - minLength; i++) {
            temp.append("0");
        }
        if (minLength == number1.length()) {
            number1 = temp + number1;
        } else number2 = temp + number2;

        for (int i = maxLength-1; i >=0; i--) {
            int tempA =Integer.parseInt(String.valueOf(number1.charAt(i)));
            int tempB = Integer.parseInt(String.valueOf(number2.charAt(i)));
            int tempI = 0;
            if (tempA + tempB >= 10) {
                tempI = tempA + tempB - 10 + flag;
                flag = 0;
                flag++;
            } else {
                tempI = tempA + tempB + flag;
                flag = 0;
            }
            result = tempI + result;
        }
        if (flag == 1)
            result = flag + result;
        return result;
    }
    public static void main(String[] args) {
        BigNumInteger bigNumInteger = new BigNumInteger();
        System.out.println(bigNumInteger.bigNumAdd(bigNumInteger.number1, bigNumInteger.number2));
    }
}