Merge pull request '旋转矩阵' (#6) from dev into master

Reviewed-on: #6

src/main/java/com/leetcode/array/RotateMatrix.java

@@ -0,0 +1,77 @@
+package com.leetcode.array;
+
+/**
+ * 
+ * 旋转矩阵
+给你一幅由 N × N 矩阵表示的图像，其中每个像素的大小为 4 字节。请你设计一种算法，将图像旋转 90 度。
+
+不占用额外内存空间能否做到？
+
+ 
+
+示例 1:
+
+给定 matrix = 
+[
+  [1,2,3],
+  [4,5,6],
+  [7,8,9]
+],
+
+原地旋转输入矩阵，使其变为:
+[
+  [7,4,1],
+  [8,5,2],
+  [9,6,3]
+]
+示例 2:
+
+给定 matrix =
+[
+  [ 5, 1, 9,11],
+  [ 2, 4, 8,10],
+  [13, 3, 6, 7],
+  [15,14,12,16]
+], 
+
+原地旋转输入矩阵，使其变为:
+[
+  [15,13, 2, 5],
+  [14, 3, 4, 1],
+  [12, 6, 8, 9],
+  [16, 7,10,11]
+]
+
+作者：力扣 (LeetCode)
+链接：https://leetcode-cn.com/leetbook/read/array-and-string/clpgd/
+来源：力扣（LeetCode）
+著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
+
+答案解析：https://leetcode-cn.com/problems/rotate-matrix-lcci/solution/xuan-zhuan-ju-zhen-by-leetcode-solution/
+ */
+public class RotateMatrix {
+
+    public void rotate(int[][] matrix) {
+        int n = matrix.length;
+        // 水平翻转
+        for (int i = 0; i < n / 2; ++i) {
+            for (int j = 0; j < n; ++j) {
+                int temp = matrix[i][j];
+                matrix[i][j] = matrix[n - i - 1][j];
+                matrix[n - i - 1][j] = temp;
+            }
+        }
+        // 主对角线翻转
+        for (int i = 0; i < n; ++i) {
+            for (int j = 0; j < i; ++j) {
+                int temp = matrix[i][j];
+                matrix[i][j] = matrix[j][i];
+                matrix[j][i] = temp;
+            }
+        }
+    }
+
+    public static void main(String[] args) {
+        
+    }
+}