415. Add Strings | 大整数相加

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

Note:

1. The length of both num1 and num2 is < 5100.
2. Both num1 and num2 contains only digits 0-9.
3. Both num1 and num2 does not contain any leading zero.
4. You must not use any built-in BigInteger library or convert the inputs to integer directly.

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思路，一个进位记录每位的进位，计算每一位的数并加在result后面，最后再逆置result即为结果，

public class Solution {
    	// 长整数相加
	public String addStrings(String num1, String num2) {
		StringBuilder result =new StringBuilder();
		int jinwei = 0;
		if (num1.length() < num2.length()) {
			String temp = num1;
			num1 = num2;
			num2 = temp;
		}
		int i = num1.length() - 1;
		int j = num2.length() - 1;
		while (j >= 0) {
			result.append((num1.charAt(i) - '0' + num2.charAt(j) - '0' + jinwei) % 10);
			jinwei = (num1.charAt(i) - '0' + num2.charAt(j) - '0' + jinwei) / 10;
			i--;
			j--;
		}
		if (jinwei > 0) {
			while (i >= 0) {
				result.append((num1.charAt(i) - '0' + jinwei) % 10);
				jinwei = (num1.charAt(i) - '0' + jinwei) / 10;
				i--;
			}
		} else {
			while (i >= 0) {
				result.append(num1.charAt(i));
				i--;
			}
		}
		if (jinwei > 0) {
			result.append(jinwei);
		}
	
		return 	result.reverse().toString();
	}
}

之前是先用String result,然后再在计算时result加上每位，再最后用新建一个stringbuilder对象用于逆置字符串，但是这样耗费时间长好多呀。

诶，写了那么多，人家用8行就写完了:

public class Solution {
    public String addStrings(String num1, String num2) {
        StringBuilder sb = new StringBuilder();
        int carry = 0;
        for(int i = num1.length() - 1, j = num2.length() - 1; i >= 0 || j >= 0 || carry == 1; i--, j--){
            int x = i < 0 ? 0 : num1.charAt(i) - '0';
            int y = j < 0 ? 0 : num2.charAt(j) - '0';
            sb.append((x + y + carry) % 10);
            carry = (x + y + carry) / 10;
        }
        return sb.reverse().toString();
    }
}

代码简洁性还加强呀！！！