537. Complex Number Multiplication
2017-04-02 19:47
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这道题主要是实现复数的乘法(主要是实现题目的特殊要求)
利用string.split(“\+”)将字符串“+”分开得到复数的实部和虚部
(a1+a2*i)*(b1+b2*i)=(a1*b1-a2*b2)+(a1*b2+a2*b1)*i,但是当结果为负数时候,有特殊的形式“+-”,所以需数要进行判断正负,具体代码如下:
利用string.split(“\+”)将字符串“+”分开得到复数的实部和虚部
(a1+a2*i)*(b1+b2*i)=(a1*b1-a2*b2)+(a1*b2+a2*b1)*i,但是当结果为负数时候,有特殊的形式“+-”,所以需数要进行判断正负,具体代码如下:
public String complexNumberMultiply(String a, String b) { String[] A=a.split("\\+"); String[] B=b.split("\\+"); int a1=Integer.parseInt(A[0]); int a2=Integer.parseInt(A[1].replace("i", "")); int b1=Integer.parseInt(B[0]); int b2=Integer.parseInt(B[1].replace("i", "")); int re=a1*b1-a2*b2; int im=a1*b2+a2*b1; StringBuffer sb=new StringBuffer(); sb.append(re); sb.append("+"); if (im<0) { sb.append("-"+Math.abs(im)); }else { sb.append(Math.abs(im)); } sb.append("i"); return sb.toString(); }
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