537. Complex Number Multiplication

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:

1. The input strings will not have extra blank.
2. The input strings will be given in the form of a+bi, where the integera andb will both belong to the range of [-100, 100]. Andthe 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();}