[Leetcode] 537. Complex Number Multiplication 解题报告
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题目:
Given two strings representing two complex numbers.
You need to return a string representing their multiplication. Note i2 = -1 according to the definition.
Example 1:
Input: "1+1i", "1+1i"Output: "0+2i"Explanation: (1 + i) * (1 + i) = 1 + i2 + 2 * i = 2i, and you need convert it to the form of 0+2i.
Example 2:
Input: "1+-1i", "1+-1i"Output: "0+-2i"Explanation: (1 - i) * (1 - i) = 1 + i2 - 2 * i = -2i, and you need convert it to the form of 0+-2i.
Note:
- The input strings will not have extra blank.
- The input strings will be given in the form of a+bi, where the integer a and b will both belong to the range of [-100, 100]. And the output should be also in this form.
思路:
哈哈,easy level的题目,思路上无需解释了。但是采用stringstream之后,代码变得出奇的简洁,请参考。
代码:
class Solution {public: string complexNumberMultiply(string a, string b) { int r1, i1, r2, i2, r, i; getComplexNumber(a, r1, i1); getComplexNumber(b, r2, i2); r = r1 * r2 - i1 * i2; i = r1 * i2 + r2 * i1; return to_string(r) + '+' + to_string(i) + 'i'; }private: void getComplexNumber(const string &s, int &r, int &i) { r = 0, i = 0; int index = 0; int real_sign = 1, imag_sign = 1; if (s[index] == '-') { real_sign = -1, ++index; } while (index < s.length() && isdigit(s[index])) { r = 10 * r + s[index++] - '0'; } ++index; // skip '+'; if (s[index] == '-') { imag_sign = -1, ++index; } while (index < s.length() && isdigit(s[index])) { i = 10 * i + s[index++] - '0'; } r *= real_sign, i *= imag_sign; return; }};
采用stringstream版本:
class Solution {public: 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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