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415. Add Strings

Description

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

You must solve the problem without using any built-in library for handling large integers (such as BigInteger). You must also not convert the inputs to integers directly.

 

Example 1:

Input: num1 = "11", num2 = "123"
Output: "134"

Example 2:

Input: num1 = "456", num2 = "77"
Output: "533"

Example 3:

Input: num1 = "0", num2 = "0"
Output: "0"

 

Constraints:

  • 1 <= num1.length, num2.length <= 104
  • num1 and num2 consist of only digits.
  • num1 and num2 don't have any leading zeros except for the zero itself.

Solutions

Solution 1: Two Pointers

We use two pointers $i$ and $j$ to point to the end of the two strings respectively, and start adding bit by bit from the end. Each time we take out the corresponding digits $a$ and $b$, calculate their sum $a + b + c$, where $c$ represents the carry from the last addition. Finally, we append the units digit of $a + b + c$ to the end of the answer string, and then take the tens digit of $a + b + c$ as the value of the carry $c$, and loop this process until the pointers of both strings have pointed to the beginning of the string and the value of the carry $c$ is $0$.

Finally, reverse the answer string and return it.

The time complexity is $O(\max(m, n))$, where $m$ and $n$ are the lengths of the two strings respectively. Ignoring the space consumption of the answer string, the space complexity is $O(1)$.

The following code also implements string subtraction, refer to the subStrings(num1, num2) function.

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class Solution:
    def addStrings(self, num1: str, num2: str) -> str:
        i, j = len(num1) - 1, len(num2) - 1
        ans = []
        c = 0
        while i >= 0 or j >= 0 or c:
            a = 0 if i < 0 else int(num1[i])
            b = 0 if j < 0 else int(num2[j])
            c, v = divmod(a + b + c, 10)
            ans.append(str(v))
            i, j = i - 1, j - 1
        return "".join(ans[::-1])

    def subStrings(self, num1: str, num2: str) -> str:
        m, n = len(num1), len(num2)
        neg = m < n or (m == n and num1 < num2)
        if neg:
            num1, num2 = num2, num1
        i, j = len(num1) - 1, len(num2) - 1
        ans = []
        c = 0
        while i >= 0:
            c = int(num1[i]) - c - (0 if j < 0 else int(num2[j]))
            ans.append(str((c + 10) % 10))
            c = 1 if c < 0 else 0
            i, j = i - 1, j - 1
        while len(ans) > 1 and ans[-1] == '0':
            ans.pop()
        if neg:
            ans.append('-')
        return ''.join(ans[::-1])

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

    public String subStrings(String num1, String num2) {
        int m = num1.length(), n = num2.length();
        boolean neg = m < n || (m == n && num1.compareTo(num2) < 0);
        if (neg) {
            String t = num1;
            num1 = num2;
            num2 = t;
        }
        int i = num1.length() - 1, j = num2.length() - 1;
        StringBuilder ans = new StringBuilder();
        for (int c