[leetcode]-537 Complex Number Multiplication

た入场券 2022-06-18 08:07 150阅读 0赞

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这道题是复数相乘的题目，个人感觉还是比较简单的。写了几个小函数解决。

intergerToString：整数转换成string类型

getRealAndImag：从string类型的复数里提取出实部和虚部（int类型）

class Solution {
    public:
        string complexNumberMultiply(string a, string b) {
            string res;
            int real1,real2,img1,img2,real,img;
            getRealAndIma(a, real1, img1);
            getRealAndIma(b,real2,img2);
            real = real1 * real2 - img1 * img2;
            img = real1 * img2 + real2 * img1;
            res = formComplexString(real, img);
            return res;
        }
        string formComplexString(int real, int img){
            string s;
            s.append(intergerToString(real));
            s.append("+");
            s.append(intergerToString(img));
            s.append("i");
            return s;
        }
        string intergerToString(int n){
            string s;
            if(n < 0){
                n = (-1) * n;
                s.append("-");
            }
            int m = n, ncount = 1;
            while(m/10){
                m = m / 10;
                ncount *= 10;
            }
            do{
                int c = n / ncount;
                string t;
                t.resize(1);
                t[0] = '0' + c;
                s.append(t);
                n = n % ncount;
                ncount /= 10;
            }while(ncount);
            return s;
        }
        void getRealAndIma(string s, int &real, int &img){
            real = 0;
            img = 0;
            int i = 0,j = 0,sign = 1;
            for(i = 0; i < s.size() && s[i] != '+'; ++ i){
                if(s[i] == '-'){
                    sign = -1;
                }else{
                    real = real * 10 + (s[i] - '0');
                }
            }
            real *= sign;
            sign = 1;
            for(j = i + 1; j < s.size() && s[j] != 'i'; ++ j){
                if(s[j] == '-'){
                    sign = -1;
                }else{
                    img = img * 10 + (s[j] - '0');
                }
            }
            img *= sign;
        }
    };

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