feat: add solutions to lc problem: No.0537

No.0537.Complex Number Multiplication

Author: yanglbme

solution/0500-0599/0537.Complex Number Multiplication/README.md

@@ -6,53 +6,108 @@
 
 <!-- 这里写题目描述 -->
 
-<p>给定两个表示<a href="https://baike.baidu.com/item/%E5%A4%8D%E6%95%B0/254365?fr=aladdin">复数</a>的字符串。</p>
+<p><a href="https://baike.baidu.com/item/%E5%A4%8D%E6%95%B0/254365?fr=aladdin" target="_blank">复数</a> 可以用字符串表示，遵循 <code>"<strong>实部</strong>+<strong>虚部</strong>i"</code> 的形式，并满足下述条件：</p>
 
-<p>返回表示它们乘积的字符串。注意，根据定义 i<sup>2</sup> = -1 。</p>
+<ul>
+	<li><code>实部</code> 是一个整数，取值范围是 <code>[-100, 100]</code></li>
+	<li><code>虚部</code> 也是一个整数，取值范围是 <code>[-100, 100]</code></li>
+	<li><code>i<sup>2</sup> == -1</code></li>
+</ul>
 
-<p><strong>示例 1:</strong></p>
+<p>给你两个字符串表示的复数 <code>num1</code> 和 <code>num2</code> ，请你遵循复数表示形式，返回表示它们乘积的字符串。</p>
+
+<p> </p>
+
+<p><strong>示例 1：</strong></p>
 
 <pre>
-<strong>输入:</strong> &quot;1+1i&quot;, &quot;1+1i&quot;
-<strong>输出:</strong> &quot;0+2i&quot;
-<strong>解释:</strong> (1 + i) * (1 + i) = 1 + i<sup>2</sup> + 2 * i = 2i ，你需要将它转换为 0+2i 的形式。
+<strong>输入：</strong>num1 = "1+1i", num2 = "1+1i"
+<strong>输出：</strong>"0+2i"
+<strong>解释：</strong>(1 + i) * (1 + i) = 1 + i2 + 2 * i = 2i ，你需要将它转换为 0+2i 的形式。
 </pre>
 
-<p><strong>示例 2:</strong></p>
+<p><strong>示例 2：</strong></p>
 
 <pre>
-<strong>输入:</strong> &quot;1+-1i&quot;, &quot;1+-1i&quot;
-<strong>输出:</strong> &quot;0+-2i&quot;
-<strong>解释:</strong> (1 - i) * (1 - i) = 1 + i<sup>2</sup> - 2 * i = -2i ，你需要将它转换为 0+-2i 的形式。 
+<strong>输入：</strong>num1 = "1+-1i", num2 = "1+-1i"
+<strong>输出：</strong>"0+-2i"
+<strong>解释：</strong>(1 - i) * (1 - i) = 1 + i2 - 2 * i = -2i ，你需要将它转换为 0+-2i 的形式。 
 </pre>
 
-<p><strong>注意:</strong></p>
+<p> </p>
 
-<ol>
-	<li>输入字符串不包含额外的空格。</li>
-	<li>输入字符串将以&nbsp;<strong>a+bi</strong> 的形式给出，其中整数 <strong>a</strong> 和 <strong>b</strong> 的范围均在 [-100, 100] 之间。<strong>输出也应当符合这种形式</strong>。</li>
-</ol>
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>num1</code> 和 <code>num2</code> 都是有效的复数表示。</li>
+</ul>
 
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
 
+`(a+bi)(c+di) = ac-bd+(ad+cb)i`。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def complexNumberMultiply(self, num1: str, num2: str) -> str:
+        a, b = map(int, num1[:-1].split('+'))
+        c, d = map(int, num2[:-1].split('+'))
+        return f'{a * c - b * d}+{a * d + c * b}i'
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    public String complexNumberMultiply(String num1, String num2) {
+        String[] c1 = num1.split("\\+|i");
+        String[] c2 = num2.split("\\+|i");
+        int a = Integer.parseInt(c1[0]);
+        int b = Integer.parseInt(c1[1]);
+        int c = Integer.parseInt(c2[0]);
+        int d = Integer.parseInt(c2[1]);
+        return String.format("%d+%di", a * c - b * d, a * d + c * b);
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    string complexNumberMultiply(string num1, string num2) {
+        int a, b, c, d;
+        sscanf(num1.c_str(), "%d+%di", &a, &b);
+        sscanf(num2.c_str(), "%d+%di", &c, &d);
+        return string(to_string(a * c - b * d) + "+" + to_string(a * d + c * b) + "i");
+    }
+};
+```
 
+### **Go**
+
+```go
+func complexNumberMultiply(num1, num2 string) string {
+	parse := func(num string) (a, b int) {
+		i := strings.IndexByte(num, '+')
+		a, _ = strconv.Atoi(num[:i])
+		b, _ = strconv.Atoi(num[i+1 : len(num)-1])
+		return
+	}
+	a, b := parse(num1)
+	c, d := parse(num2)
+	return fmt.Sprintf("%d+%di", a*c-b*d, a*d+b*c)
+}
 ```
 
 ### **...**

solution/0500-0599/0537.Complex Number Multiplication/README_EN.md

@@ -4,70 +4,100 @@
 
 ## Description
 
-<p>
+<p>A <a href="https://en.wikipedia.org/wiki/Complex_number" target="_blank">complex number</a> can be represented as a string on the form <code>&quot;<strong>real</strong>+<strong>imaginary</strong>i&quot;</code> where:</p>
 
-Given two strings representing two <a href = "https://en.wikipedia.org/wiki/Complex_number">complex numbers</a>.</p>
+<ul>
+	<li><code>real</code> is the real part and is an integer in the range <code>[-100, 100]</code>.</li>
+	<li><code>imaginary</code> is the imaginary part and is an integer in the range <code>[-100, 100]</code>.</li>
+	<li><code>i<sup>2</sup> == -1</code>.</li>
+</ul>
 
-<p>
+<p>Given two complex numbers <code>num1</code> and <code>num2</code> as strings, return <em>a string of the complex number that represents their multiplications</em>.</p>
 
-You need to return a string representing their multiplication. Note i<sup>2</sup> = -1 according to the definition.
-
-</p>
-
-<p><b>Example 1:</b><br />
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
 
 <pre>
-
-<b>Input:</b> "1+1i", "1+1i"
-
-<b>Output:</b> "0+2i"
-
-<b>Explanation:</b> (1 + i) * (1 + i) = 1 + i<sup>2</sup> + 2 * i = 2i, and you need convert it to the form of 0+2i.
-
+<strong>Input:</strong> num1 = &quot;1+1i&quot;, num2 = &quot;1+1i&quot;
+<strong>Output:</strong> &quot;0+2i&quot;
+<strong>Explanation:</strong> (1 + i) * (1 + i) = 1 + i2 + 2 * i = 2i, and you need convert it to the form of 0+2i.
 </pre>
 
-</p>
-
-<p><b>Example 2:</b><br />
+<p><strong>Example 2:</strong></p>
 
 <pre>
-
-<b>Input:</b> "1+-1i", "1+-1i"
-
-<b>Output:</b> "0+-2i"
-
-<b>Explanation:</b> (1 - i) * (1 - i) = 1 + i<sup>2</sup> - 2 * i = -2i, and you need convert it to the form of 0+-2i.
-
+<strong>Input:</strong> num1 = &quot;1+-1i&quot;, num2 = &quot;1+-1i&quot;
+<strong>Output:</strong> &quot;0+-2i&quot;
+<strong>Explanation:</strong> (1 - i) * (1 - i) = 1 + i2 - 2 * i = -2i, and you need convert it to the form of 0+-2i.
 </pre>
 
-</p>
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
 
-<p><b>Note:</b>
-
-<ol>
-
-<li>The input strings will not have extra blank.</li>
-
-<li>The input strings will be given in the form of <b>a+bi</b>, where the integer <b>a</b> and <b>b</b> will both belong to the range of [-100, 100]. And <b>the output should be also in this form</b>.</li>
-
-</ol>
-
-</p>
+<ul>
+	<li><code>num1</code> and <code>num2</code> are valid complex numbers.</li>
+</ul>
 
 ## Solutions
 
+`(a+bi)(c+di) = ac-bd+(ad+cb)i`
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 ```python
-
+class Solution:
+    def complexNumberMultiply(self, num1: str, num2: str) -> str:
+        a, b = map(int, num1[:-1].split('+'))
+        c, d = map(int, num2[:-1].split('+'))
+        return f'{a * c - b * d}+{a * d + c * b}i'
 ```
 
 ### **Java**
 
 ```java
+class Solution {
+    public String complexNumberMultiply(String num1, String num2) {
+        String[] c1 = num1.split("\\+|i");
+        String[] c2 = num2.split("\\+|i");
+        int a = Integer.parseInt(c1[0]);
+        int b = Integer.parseInt(c1[1]);
+        int c = Integer.parseInt(c2[0]);
+        int d = Integer.parseInt(c2[1]);
+        return String.format("%d+%di", a * c - b * d, a * d + c * b);
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    string complexNumberMultiply(string num1, string num2) {
+        int a, b, c, d;
+        sscanf(num1.c_str(), "%d+%di", &a, &b);
+        sscanf(num2.c_str(), "%d+%di", &c, &d);
+        return string(to_string(a * c - b * d) + "+" + to_string(a * d + c * b) + "i");
+    }
+};
+```
 
+### **Go**
+
+```go
+func complexNumberMultiply(num1, num2 string) string {
+	parse := func(num string) (a, b int) {
+		i := strings.IndexByte(num, '+')
+		a, _ = strconv.Atoi(num[:i])
+		b, _ = strconv.Atoi(num[i+1 : len(num)-1])
+		return
+	}
+	a, b := parse(num1)
+	c, d := parse(num2)
+	return fmt.Sprintf("%d+%di", a*c-b*d, a*d+b*c)
+}
 ```
 
 ### **...**

solution/0500-0599/0537.Complex Number Multiplication/Solution.cpp

@@ -0,0 +1,9 @@
+class Solution {
+public:
+    string complexNumberMultiply(string num1, string num2) {
+        int a, b, c, d;
+        sscanf(num1.c_str(), "%d+%di", &a, &b);
+        sscanf(num2.c_str(), "%d+%di", &c, &d);
+        return string(to_string(a * c - b * d) + "+" + to_string(a * d + c * b) + "i");
+    }
+};

solution/0500-0599/0537.Complex Number Multiplication/Solution.go

@@ -0,0 +1,11 @@
+func complexNumberMultiply(num1, num2 string) string {
+	parse := func(num string) (a, b int) {
+		i := strings.IndexByte(num, '+')
+		a, _ = strconv.Atoi(num[:i])
+		b, _ = strconv.Atoi(num[i+1 : len(num)-1])
+		return
+	}
+	a, b := parse(num1)
+	c, d := parse(num2)
+	return fmt.Sprintf("%d+%di", a*c-b*d, a*d+b*c)
+}

solution/0500-0599/0537.Complex Number Multiplication/Solution.java

@@ -0,0 +1,11 @@
+class Solution {
+    public String complexNumberMultiply(String num1, String num2) {
+        String[] c1 = num1.split("\\+|i");
+        String[] c2 = num2.split("\\+|i");
+        int a = Integer.parseInt(c1[0]);
+        int b = Integer.parseInt(c1[1]);
+        int c = Integer.parseInt(c2[0]);
+        int d = Integer.parseInt(c2[1]);
+        return String.format("%d+%di", a * c - b * d, a * d + c * b);
+    }
+}

solution/0500-0599/0537.Complex Number Multiplication/Solution.py

@@ -0,0 +1,5 @@
+class Solution:
+    def complexNumberMultiply(self, num1: str, num2: str) -> str:
+        a, b = map(int, num1[:-1].split('+'))
+        c, d = map(int, num2[:-1].split('+'))
+        return f'{a * c - b * d}+{a * d + c * b}i'