Cinte's Leetcode Record
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415. 字符串相加

Posted on Edited on In Disqus:
Symbols count in article: 1.1k Reading time ≈ 1 mins.

415. 字符串相加

效率很低的模拟方法
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class Solution {
public:
    string addStrings(string num1, string num2) {
        int c = 0;
        int a = num1.size() - 1, b = num2.size() - 1;
        string ret;
        while (a >= 0 && b >= 0)
        {
            int i = num1[a--] + num2[b--] - '0' - '0' + c;
            c = i / 10;
            i %= 10;
            ret = char(i + '0') + ret;
        }
        while (c != 0 && (a >= 0 || b >= 0))
        {
            int i;
            if (a >= 0)
                i = num1[a--] - '0' + c;
            else
                i = num2[b--] - '0' + c;

            c = i / 10;
            i %= 10;
            ret = char(i + '0') + ret;
        }
        if (c != 0)
            ret = char(c + '0') + ret;
        if (a >= 0 || b >= 0)
            ret = a >= 0 ? num1.substr(0, a + 1) + ret : num2.substr(0, b + 1) + ret;
        return ret;
    }
};

在字符串的开头添加字符会导致每次都要重新分配字符串
为了避免多次重新分配字符串，采用先加再反转来避免，顺便可以合并其中的一部分块

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class Solution {
public:
    string addStrings(string num1, string num2) {
        int c = 0;
        int a = num1.size() - 1, b = num2.size() - 1;
        string ret;
        while (a >= 0 || b >= 0 || c)
        {
            int i = 0;
            if(a >= 0)
                i += num1[a--] - '0';
            if(b >= 0)
                i += num2[b--] - '0';
            i += c;
            c = i / 10;
            i %= 10;
            ret += char(i + '0');
        }
        reverse(ret.begin(), ret.end());
        return ret;
    }
};