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;

	}

};