537. Complex Number Multiplication-Medium
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大水题!leetcode大概是看答案代码行数判断难易程度的吧。
写一道Medium动态规划写到心态崩了,所以今天先水一道,下周再填那个坑。
Description
给定两个用string表示的复数输出乘积。
Example:
Input: “1+1i”, “1+1i”
Output: “0+2i”
Code
#include <cstdlib>#include <sstream>class Solution {public: string complexNumberMultiply(string a, string b) { pair<int, int> trsA = transform(a); pair<int, int> trsB = transform(b); int real = trsA.first*trsB.first - trsA.second*trsB.second; int imaginary = trsA.second*trsB.first + trsA.first*trsB.second; string ret; stringstream ss; ss << real << "+" << imaginary << "i"; ss >> ret; return ret; } pair<int, int> transform(string a) { size_t plus_pos = a.find('+'); int real = atoi(a.substr(0, plus_pos).c_str()); int imaginary = atoi(a.substr(plus_pos+1, a.size()-plus_pos-2).c_str()); pair<int, int> ret(real, imaginary); return ret; }};
Summary
1.string转int:用cstdlib中的atoi(string.c_str())
2.int转string:用sstream中的stringstream,造一个这种流对象,往对象中输入int,输出到目标string。同时可用该对象做拼接。
3.size_t也不过就是unsigned int,要加要减随便当int用。
下面照例放上“别人家的代码”:
stringstream玩得好
string complexNumberMultiply(string a, string b) { int ra, ia, rb, ib; char buff; stringstream aa(a), bb(b), ans; aa >> ra >> buff >> ia >> buff; bb >> rb >> buff >> ib >> buff; ans << ra*rb - ia*ib << "+" << ra*ib + rb*ia << "i"; return ans.str();}
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