feat: add solutions to lc problem: No.0836

No.0836.Rectangle Overlap

Author: yanglbme

solution/0800-0899/0836.Rectangle Overlap/README.md

@@ -50,22 +50,70 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：判断不重叠的情况**
+
+我们记矩形 $rec1$ 的坐标点为 $(x_1, y_1, x_2, y_2)$，矩形 $rec2$ 的坐标点为 $(x_3, y_3, x_4, y_4)$。
+
+那么当满足以下任一条件时，矩形 $rec1$ 和 $rec2$ 不重叠：
+
+-   满足 $y_3 \geq y_2$，即 $rec2$ 在 $rec1$ 的上方；
+-   满足 $y_4 \leq y_1$，即 $rec2$ 在 $rec1$ 的下方；
+-   满足 $x_3 \geq x_2$，即 $rec2$ 在 $rec1$ 的右方；
+-   满足 $x_4 \leq x_1$，即 $rec2$ 在 $rec1$ 的左方。
+
+当以上条件都不满足时，矩形 $rec1$ 和 $rec2$ 重叠。
+
+时间复杂度 $O(1)$，空间复杂度 $O(1)$。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def isRectangleOverlap(self, rec1: List[int], rec2: List[int]) -> bool:
+        x1, y1, x2, y2 = rec1
+        x3, y3, x4, y4 = rec2
+        return not (y3 >= y2 or y4 <= y1 or x3 >= x2 or x4 <= x1)
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    public boolean isRectangleOverlap(int[] rec1, int[] rec2) {
+        int x1 = rec1[0], y1 = rec1[1], x2 = rec1[2], y2 = rec1[3];
+        int x3 = rec2[0], y3 = rec2[1], x4 = rec2[2], y4 = rec2[3];
+        return !(y3 >= y2 || y4 <= y1 || x3 >= x2 || x4 <= x1);
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    bool isRectangleOverlap(vector<int>& rec1, vector<int>& rec2) {
+        int x1 = rec1[0], y1 = rec1[1], x2 = rec1[2], y2 = rec1[3];
+        int x3 = rec2[0], y3 = rec2[1], x4 = rec2[2], y4 = rec2[3];
+        return !(y3 >= y2 || y4 <= y1 || x3 >= x2 || x4 <= x1);
+    }
+};
+```
+
+### **Go**
 
+```go
+func isRectangleOverlap(rec1 []int, rec2 []int) bool {
+	x1, y1, x2, y2 := rec1[0], rec1[1], rec1[2], rec1[3]
+	x3, y3, x4, y4 := rec2[0], rec2[1], rec2[2], rec2[3]
+	return !(y3 >= y2 || y4 <= y1 || x3 >= x2 || x4 <= x1)
+}
 ```
 
 ### **...**

solution/0800-0899/0836.Rectangle Overlap/README_EN.md

@@ -38,13 +38,46 @@
 ### **Python3**
 
 ```python
-
+class Solution:
+    def isRectangleOverlap(self, rec1: List[int], rec2: List[int]) -> bool:
+        x1, y1, x2, y2 = rec1
+        x3, y3, x4, y4 = rec2
+        return not (y3 >= y2 or y4 <= y1 or x3 >= x2 or x4 <= x1)
 ```
 
 ### **Java**
 
 ```java
+class Solution {
+    public boolean isRectangleOverlap(int[] rec1, int[] rec2) {
+        int x1 = rec1[0], y1 = rec1[1], x2 = rec1[2], y2 = rec1[3];
+        int x3 = rec2[0], y3 = rec2[1], x4 = rec2[2], y4 = rec2[3];
+        return !(y3 >= y2 || y4 <= y1 || x3 >= x2 || x4 <= x1);
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    bool isRectangleOverlap(vector<int>& rec1, vector<int>& rec2) {
+        int x1 = rec1[0], y1 = rec1[1], x2 = rec1[2], y2 = rec1[3];
+        int x3 = rec2[0], y3 = rec2[1], x4 = rec2[2], y4 = rec2[3];
+        return !(y3 >= y2 || y4 <= y1 || x3 >= x2 || x4 <= x1);
+    }
+};
+```
+
+### **Go**
 
+```go
+func isRectangleOverlap(rec1 []int, rec2 []int) bool {
+	x1, y1, x2, y2 := rec1[0], rec1[1], rec1[2], rec1[3]
+	x3, y3, x4, y4 := rec2[0], rec2[1], rec2[2], rec2[3]
+	return !(y3 >= y2 || y4 <= y1 || x3 >= x2 || x4 <= x1)
+}
 ```
 
 ### **...**

solution/0800-0899/0836.Rectangle Overlap/Solution.cpp

@@ -1,14 +1,8 @@
 class Solution {
 public:
     bool isRectangleOverlap(vector<int>& rec1, vector<int>& rec2) {
-        int l1 = rec1[1], u1 = rec1[0], r1 = rec1[3], d1 = rec1[2];
-        int l2 = rec2[1], u2 = rec2[0], r2 = rec2[3], d2 = rec2[2];
-
-        //printf("1: (%d,%d), (%d,%d)\n", l1, u1, r1, d1) ;
-        //printf("2: (%d,%d), (%d,%d)\n", l2, u2, r2, d2) ;
-
-        if (l1 < r2 && u1 < d2 && l2 < r1 && u2 < d1)
-            return true;
-        return false;
+        int x1 = rec1[0], y1 = rec1[1], x2 = rec1[2], y2 = rec1[3];
+        int x3 = rec2[0], y3 = rec2[1], x4 = rec2[2], y4 = rec2[3];
+        return !(y3 >= y2 || y4 <= y1 || x3 >= x2 || x4 <= x1);
     }
 };

solution/0800-0899/0836.Rectangle Overlap/Solution.go

@@ -0,0 +1,5 @@
+func isRectangleOverlap(rec1 []int, rec2 []int) bool {
+	x1, y1, x2, y2 := rec1[0], rec1[1], rec1[2], rec1[3]
+	x3, y3, x4, y4 := rec2[0], rec2[1], rec2[2], rec2[3]
+	return !(y3 >= y2 || y4 <= y1 || x3 >= x2 || x4 <= x1)
+}

solution/0800-0899/0836.Rectangle Overlap/Solution.java

@@ -0,0 +1,7 @@
+class Solution {
+    public boolean isRectangleOverlap(int[] rec1, int[] rec2) {
+        int x1 = rec1[0], y1 = rec1[1], x2 = rec1[2], y2 = rec1[3];
+        int x3 = rec2[0], y3 = rec2[1], x4 = rec2[2], y4 = rec2[3];
+        return !(y3 >= y2 || y4 <= y1 || x3 >= x2 || x4 <= x1);
+    }
+}

solution/0800-0899/0836.Rectangle Overlap/Solution.py

@@ -0,0 +1,5 @@
+class Solution:
+    def isRectangleOverlap(self, rec1: List[int], rec2: List[int]) -> bool:
+        x1, y1, x2, y2 = rec1
+        x3, y3, x4, y4 = rec2
+        return not (y3 >= y2 or y4 <= y1 or x3 >= x2 or x4 <= x1)

solution/1000-1099/1053.Previous Permutation With One Swap/README.md

@@ -51,9 +51,9 @@
 
 **方法一：贪心**
 
-我们先从右到左遍历数组，找到第一个满足 `arr[i - 1] > arr[i]` 的下标 `i`，此时 `arr[i - 1]` 就是我们要交换的数字，我们再从右到左遍历数组，找到第一个满足 `arr[j] < arr[i - 1]` 且 `arr[j] != arr[j - 1]` 的下标 `j`，此时我们交换 `arr[i - 1]` 和 `arr[j]` 后返回即可。
+我们先从右到左遍历数组，找到第一个满足 $arr[i - 1] \gt arr[i]$ 的下标 $i$，此时 $arr[i - 1]$ 就是我们要交换的数字，我们再从右到左遍历数组，找到第一个满足 $arr[j] \lt arr[i - 1]$ 且 $arr[j] \neq arr[j - 1]$ 的下标 $j$，此时我们交换 $arr[i - 1]$ 和 $arr[j]$ 后返回即可。
 
-如果遍历完数组都没有找到满足条件的下标 `i`，说明数组已经是最小排列，直接返回原数组即可。
+如果遍历完数组都没有找到满足条件的下标 $i$，说明数组已经是最小排列，直接返回原数组即可。
 
 时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组长度。
 

solution/1000-1099/1054.Distant Barcodes/README.md

@@ -44,7 +44,7 @@
 
 重排条形码时，我们每次从堆顶弹出一个元素 `(v, k)`，将 `k` 添加到结果数组中，并将 `(v-1, k)` 放入队列 `q` 中。当队列长度大于 $1$ 时，弹出队首元素 `p`，若此时 `p.v` 大于 $0$，则将 `p` 放入堆中。循环，直至堆为空。
 
-时间复杂度 $O(n\log n)$，空间复杂度 $O(n)$。其中 $n$ 为条形码数组的长度。
+时间复杂度 $O(n \times \log n)$，空间复杂度 $O(n)$。其中 $n$ 为条形码数组的长度。
 
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