【数据结构基础_字符串】Leetcode 415.字符串相加

原题链接：Leetcode 415. Add Strings

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"
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Example 2:

Input: num1 = "456", num2 = "77"
Output: "533"
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Example 3:

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

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方法一：模拟

思路：

模拟竖式加法，用cf表示进位
问题是当两个数的位数不一样怎么办？
解决办法，允许下标到负数，给0就行。这样相当于加上了先导0，实现了两个数的位数相等

c++代码：

class Solution {
public:
    string addStrings(string num1, string num2) {
        // ij为两个指针, 初始指向末尾 cf为进位标志
        int i = num1.size() - 1, j = num2.size() - 1, cf = 0;
        string ans = "";

        while(i >= 0 || j >= 0 || cf != 0){
            int x = (i >= 0) ? num1[i] - '0' : 0;
            int y = (j >= 0) ? num2[j] - '0' : 0;

            // cnt当前位的和
            int cnt = x + y + cf;

            ans += ('0' + cnt % 10);
            cf = cnt / 10;
            i--;
            j--;
        }

        // 翻转
        reverse(ans.begin(), ans.end());

        return ans;
    }
};
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复杂度分析：

• 时间复杂度：O(max(n1, n2))，运算的次数取决于两个数中较大的那个数的位数
• 空间复杂度：O(1)，常数个临时变量