537. Complex Number Multiplication
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.这个题思路很简单，就是实现复数的乘法，但是由于输入输出都是string所以转换比较麻烦，最开始我是想通过写一个toInt函数来将字符串转化为数字来做，可以通过但是不够简洁。
class Solution { public: int toint(string t) { int f = 0; if (t[0] == '-') f = 1; int c = 0; for (int i = t.size()-1; i >= f; i--) { c = c + (t[i] - 48)*pow(10, t.size() - 1 - i); } return f ? -c : c; } string complexNumberMultiply(string a, string b) { string ap, ab, bp, bb; int i, j; for (i = 0; a[i] != '+'; i++) ap = ap + a[i]; for (j = 0; b[j] != '+'; j++) bp = bp + b[j]; for (i = i + 1; a[i] != 'i'; i++) ab = ab + a[i]; for (j = j + 1; b[j] != 'i'; j++) bb = bb + b[j]; int a1, a2, b1, b2; a1 = toint(ap); a2 = toint(ab); b1 = toint(bp); b2 = toint(bb); return to_string(a1*b1 - a2*b2) + "+" + to_string(a1*b2 + a2*b1) + "i"; } };
大神通过stringstream来做就相当简洁易懂。
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(); }
