Leetcode: Multiply Strings

题目

大整数乘法

 

思路:

1. 转化成分组背包问题, 代码比较 tricky

2. 将 string 相乘转化为 string 按位乘再加的过程

3. 细节. 不同长度 string 相加可以先补 0, 对齐. 按位相乘的时候, 可以从左向右计算, 便于移位

 

代码:

class Solution {
public:
    string multiply(string num1, string num2) {
		if(num1.size() <= 0 || num2.size() <= 0)
			return "";
		if(num1.size() == 1 && num1[0] == '0')
			return "0";
		if(num2.size() == 1 && num2[0] == '0')
			return "0";

		if(num1.size() < num2.size()) {
			swap(num1, num2);
		}
		string res;
		for(int i = 0; i < num2.size(); i++) {
			string party = oneBitMultipy(num1, num2[i]-'0');
			res.push_back('0');
			res = add(res, party);
		}
		return res;
    }

	string add(string num1, string num2) {
		string res = "";
		int len1 = num1.size(), len2 = num2.size();
		if(len1 < len2) {
			res.append(num2.size()-num1.size(), '0').append(num1);
			num1 = res;
		}else if(len1 > len2) {
			res.append(num1.size()-num2.size(), '0').append(num2);
			num2 = res;
		}

		res.clear();
		int leftover = 0, len = num1.size()-1;
		int i;
		for(i = 0; i < num1.size(); i ++) {
			int tmp = leftover+num1[len-i]-'0'+num2[len-i]-'0';
			res.push_back(tmp%10+'0');
			leftover = tmp/10;
		}
		
		if(leftover) {
			res.push_back('0'+leftover);
		}
		reverse(res.begin(), res.end());
		return res;
	}

	string oneBitMultipy(string num1, int bit) {
		string res;
		int leftover = 0;
		for(int i = num1.size()-1; i >= 0; i--) {
			int tmp = (num1[i]-'0')*bit + leftover;
			res.push_back(tmp%10+'0');
			leftover = tmp/10;
		}
		if(leftover) {
			res.push_back(leftover+'0');
		}
		reverse(res.begin(), res.end());
		return res;
	}
};