[LeetCode] 415. Add Strings 字符串相加

Given two non-negative numbers 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.

两个只含有数字的字符串相加。

解法: 对于每个字符转成对应的整数，然后相加，结果在写入res。

Java:

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();
    }
}　　

Python:

class Solution(object):
    def addStrings(self, num1, num2):
        """
        :type num1: str
        :type num2: str
        :rtype: str
        """
        result = []
        i, j, carry = len(num1) - 1, len(num2) - 1, 0
        
        while i >= 0 or j >= 0 or carry:
            if i >= 0:
                carry += ord(num1[i]) - ord('0');
                i -= 1
            if j >= 0:
                carry += ord(num2[j]) - ord('0');
                j -= 1
            result.append(str(carry % 10))
            carry /= 10
        result.reverse()

        return "".join(result)

C++:

class Solution {
public:
    string addStrings(string num1, string num2) {
        string res = "";
        int m = num1.size(), n = num2.size(), i = m - 1, j = n - 1, carry = 0;
        while (i >= 0 || j >= 0) {
            int a = i >= 0 ? num1[i--] - '0' : 0;
            int b = j >= 0 ? num2[j--] - '0' : 0;
            int sum = a + b + carry;
            res.insert(res.begin(), sum % 10 + '0');
            carry = sum / 10;
        }
        return carry ? "1" + res : res;
    }
};

　　　　

 

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转载于:https://www.cnblogs.com/lightwindy/p/8648835.html