[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:

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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:

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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++:

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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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