LeetCode[415] Add Strings

最新推荐文章于 2025-08-19 21:22:03 发布

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最新推荐文章于 2025-08-19 21:22:03 发布

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文章标签： leetcode string

本文链接：https://blog.youkuaiyun.com/wrsophia_977/article/details/52785902

本文介绍了一种不使用内置大数库或直接转换为整数的方法来实现两个非负数字符串的相加操作。通过反转字符串并逐位计算的方式避免了大数运算的复杂性。

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

Note:

The length of both num1 and num2 is < 5100.
Both num1 and num2 contains only digits 0-9.
Both num1 and num2 does not contain any leading zero.
You must not use any built-in BigInteger library or convert the inputs to integer directly.

先将num1和num2翻转，最后再将ans翻转。需注意最后可能需要加上进位。

class Solution {
public:
	string addStrings(string num1, string num2) {
		string ans = "";
		int flag = 0;
		reverse(num1.begin(), num1.end());
		reverse(num2.begin(), num2.end());
		int dig1 = 0, dig2 = 0;
		auto it1 = num1.begin(), it2 = num2.begin();
		while (it1 != num1.end() && it2 != num2.end()) {
			int tmp = *it1 - '0' + *it2 - '0' + flag;
			flag = tmp / 10;
			tmp %= 10;
			ans.push_back(tmp + '0');
			++it1, ++it2;
		}
		while (it1 != num1.end()) {
			int tmp = *it1 - '0' + flag;
			flag = tmp / 10;
			tmp %= 10;
			ans.push_back(tmp + '0');
			++it1;
		}
		while (it2 != num2.end()) {
			int tmp = *it2 - '0' + flag;
			flag = tmp / 10;
			tmp %= 10;
			ans.push_back(tmp + '0');
			++it2;
		}
		if (flag == 1)
		    ans += '1';
		reverse(ans.begin(), ans.end());
		return ans;
	}
};