feat: add solutions to lc problem: No.0223

No.0223.Rectangle Area

Author: yanglbme

solution/0200-0299/0223.Rectangle Area/README.md

@@ -45,7 +45,11 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
-计算重叠部分的面积，注意考虑没有重叠的情况
+**方法一：计算重叠面积**
+
+我们先计算出两个矩形各自的面积，记为 $a$ 和 $b$，然后计算重叠的宽度 $width$ 和高度 $height$，那么重叠的面积为 $max(width, 0) \times max(height, 0)$，最后将 $a$、$b$ 和重叠面积相减即可。
+
+时间复杂度 $O(1)$，空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -130,6 +134,41 @@ func min(a, b int) int {
 }
 ```
 
+### **TypeScript**
+
+```ts
+function computeArea(
+    ax1: number,
+    ay1: number,
+    ax2: number,
+    ay2: number,
+    bx1: number,
+    by1: number,
+    bx2: number,
+    by2: number,
+): number {
+    const a = (ax2 - ax1) * (ay2 - ay1);
+    const b = (bx2 - bx1) * (by2 - by1);
+    const width = Math.min(ax2, bx2) - Math.max(ax1, bx1);
+    const height = Math.min(ay2, by2) - Math.max(ay1, by1);
+    return a + b - Math.max(width, 0) * Math.max(height, 0);
+}
+```
+
+### **C#**
+
+```cs
+public class Solution {
+    public int ComputeArea(int ax1, int ay1, int ax2, int ay2, int bx1, int by1, int bx2, int by2) {
+        int a = (ax2 - ax1) * (ay2 - ay1);
+        int b = (bx2 - bx1) * (by2 - by1);
+        int width = Math.Min(ax2, bx2) - Math.Max(ax1, bx1);
+        int height = Math.Min(ay2, by2) - Math.Max(ay1, by1);
+        return a + b - Math.Max(height, 0) * Math.Max(width, 0);
+    }
+}
+```
+
 ### **...**
 
 ```

solution/0200-0299/0223.Rectangle Area/README_EN.md

@@ -116,6 +116,41 @@ func min(a, b int) int {
 }
 ```
 
+### **TypeScript**
+
+```ts
+function computeArea(
+    ax1: number,
+    ay1: number,
+    ax2: number,
+    ay2: number,
+    bx1: number,
+    by1: number,
+    bx2: number,
+    by2: number,
+): number {
+    const a = (ax2 - ax1) * (ay2 - ay1);
+    const b = (bx2 - bx1) * (by2 - by1);
+    const width = Math.min(ax2, bx2) - Math.max(ax1, bx1);
+    const height = Math.min(ay2, by2) - Math.max(ay1, by1);
+    return a + b - Math.max(width, 0) * Math.max(height, 0);
+}
+```
+
+### **C#**
+
+```cs
+public class Solution {
+    public int ComputeArea(int ax1, int ay1, int ax2, int ay2, int bx1, int by1, int bx2, int by2) {
+        int a = (ax2 - ax1) * (ay2 - ay1);
+        int b = (bx2 - bx1) * (by2 - by1);
+        int width = Math.Min(ax2, bx2) - Math.Max(ax1, bx1);
+        int height = Math.Min(ay2, by2) - Math.Max(ay1, by1);
+        return a + b - Math.Max(height, 0) * Math.Max(width, 0);
+    }
+}
+```
+
 ### **...**
 
 ```

solution/0200-0299/0223.Rectangle Area/Solution.cs

@@ -1,20 +1,9 @@
-// https://leetcode.com/problems/rectangle-area/
-
-using System;
-
-public partial class Solution
-{
-    public int ComputeArea(int A, int B, int C, int D, int E, int F, int G, int H)
-    {
-        var area = (long)(C - A) * (D - B) + (G - E) * (H - F);
-        var overlapA = Math.Max(A, E);
-        var overlapB = Math.Max(B, F);
-        var overlapC = Math.Min(C, G);
-        var overlapD = Math.Min(D, H);
-        if (overlapA < overlapC && overlapB < overlapD)
-        {
-            area -= (overlapC - overlapA) * (overlapD - overlapB);
-        }
-        return (int)area;
+public class Solution {
+    public int ComputeArea(int ax1, int ay1, int ax2, int ay2, int bx1, int by1, int bx2, int by2) {
+        int a = (ax2 - ax1) * (ay2 - ay1);
+        int b = (bx2 - bx1) * (by2 - by1);
+        int width = Math.Min(ax2, bx2) - Math.Max(ax1, bx1);
+        int height = Math.Min(ay2, by2) - Math.Max(ay1, by1);
+        return a + b - Math.Max(height, 0) * Math.Max(width, 0);
     }
 }

solution/0200-0299/0223.Rectangle Area/Solution.ts

@@ -0,0 +1,16 @@
+function computeArea(
+    ax1: number,
+    ay1: number,
+    ax2: number,
+    ay2: number,
+    bx1: number,
+    by1: number,
+    bx2: number,
+    by2: number,
+): number {
+    const a = (ax2 - ax1) * (ay2 - ay1);
+    const b = (bx2 - bx1) * (by2 - by1);
+    const width = Math.min(ax2, bx2) - Math.max(ax1, bx1);
+    const height = Math.min(ay2, by2) - Math.max(ay1, by1);
+    return a + b - Math.max(width, 0) * Math.max(height, 0);
+}