LeetCode Weekly Contest 25 之 537.Complex Number Multiplication

LeetCode Weekly Contest 25

赛题

本次周赛主要分为以下4道题：

507 Perfect Number (3分)

537 Complex Number Multiplication (6分)

545 boundary of Binary Tree (8分)

546 Remove Boxes (9分)

537 Complex Number Multiplication

Problem:

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.

依旧没有难度，这道题考的是对字符串的处理，但我又把它想复杂了，浪费了大量时间在解析上，就针对我的解题思路，来慢慢优化。

My first solution(6ms)

public String complexNumberMultiply(String a, String b) {

int a1 = 0;
int a2 = 0;

StringBuilder a1Builder = new StringBuilder();
StringBuilder a2Builder = new StringBuilder();
for(int i = 0; i < a.length(); i++){
if(a.charAt(i) != '+')
a1Builder.append(a.charAt(i));
if(a.charAt(i) == '+'){
a2Builder.append(a.substring(i+1,a.length()-1));
break;
}
}

a1 = Integer.parseInt(a1Builder.toString());
a2 = Integer.parseInt(a2Builder.toString());

int b1 = 0;
int b2 = 0;

StringBuilder b1Builder = new StringBuilder();
StringBuilder b2Builder = new StringBuilder();
for(int i = 0; i < b.length(); i++){
if(b.charAt(i) != '+')
b1Builder.append(b.charAt(i));

if(b.charAt(i) == '+'){
b2Builder.append(b.substring(i+1,b.length()-1));
break;
}
}

b1 = Integer.parseInt(b1Builder.toString());
b2 = Integer.parseInt(b2Builder.toString());

StringBuilder res = new StringBuilder();

int c1 = a1*b1 - a2*b2;
int c2 = a1 * b2 + a2 * b1;

res.append(String.valueOf(c1));
res.append("+");
res.append(String.valueOf(c2));
res.append("i");

return res.toString();
}

上述代码有个细节，刚开始我在code时，还区分了负号，但

String.valueOf()

是支持对负号提取的，所以无须多此一举。

该代码有大量重复的内容，如for循环，对“+”号的分解。这一操作用java的split方法就可以轻松实现了。其实核心思想个是加法的二元操作。一次split，就能区分加法的左和右，所以完善后的代码为：

My second solution(12ms)

public String complexNumberMultiply(String a, String b) {

int a1 = Integer.parseInt(a.split("\\+|i")[0]);
int a2 = Integer.parseInt(a.split("\\+|i")[1]);

int b1 = Integer.parseInt(b.split("\\+|i")[0]);
int b2 = Integer.parseInt(b.split("\\+|i")[1]);

String res = "";

int c1 = a1 * b1 - a2 * b2;
int c2 = a1 * b2 + a2 * b1;

res = String.valueOf(c1)+"+"+String.valueOf(c2)+"i";

return res;
}

代码简洁很多，用到了split方法和正则表达式，其中加号的描述需要转义，且在解析虚数位置时，要去除符号

i

。实际的运行效率比较低，且并不知道是如何对字符串做分解的，所以我们用一种更加效率的拆分，来实现它。

My third solution(5ms)

public String complexNumberMultiply(String a, String b) {
int[] aa = parseComplex(a);
int[] bb = parseComplex(b);

int c1 = aa[0] * bb[0] - aa[1] * bb[1];
int c2 = aa[0] * bb[1] + aa[1] * bb[0];

return String.valueOf(c1)+"+"+String.valueOf(c2)+"i";
}

private int[] parseComplex(String complex){

int[] res = new int[2];

int i = 0;
for(; complex.charAt(i) != '+'; i++);

res[0] = Integer.parseInt(complex.substring(0,i));
res[1] = Integer.parseInt(complex.substring(i+1,complex.length()-1));

return res;
}

这种查找加法的方式，的确头一次见，但很管用，

i

必须是个全局变量，它的思路很简单，找到“+”号所在的位置，有了

i

，就能对字符串做切割了，不需要向我第一种解决方案，那么复杂，还得append左半部分。哇，啥时候能写出如此优美的代码啊！