refactor 498

Author: fishercoder1534

src/main/java/com/fishercoder/solutions/_498.java

@@ -1,71 +1,50 @@
 package com.fishercoder.solutions;
 
-/**
- * 498. Diagonal Traverse
- *
- *  Given a matrix of m x N elements (m rows, N columns), return all elements of the matrix in diagonal order
- *  as shown in the below image.
-
- Example:
-
- Input:
- [
- [ 1, 2, 3 ],
- [ 4, 5, 6 ],
- [ 7, 8, 9 ]
- ]
- Output:  [1,2,4,7,5,3,6,8,9]
-
- Note:
-
- The total number of elements of the given matrix will not exceed 10,000.
-
- */
 public class _498 {
 
-public static class Solutoin1 {
-    /**
-     * Reference: https://discuss.leetcode.com/topic/77865/concise-java-solution/2
-     * Just keep walking the matrix, when hitting the four borders (top, bottom, left or right),
-     * just directions and keep walking.
-     */
-    public int[] findDiagonalOrder(int[][] matrix) {
+    public static class Solutoin1 {
+        /**
+         * Reference: https://discuss.leetcode.com/topic/77865/concise-java-solution/2
+         * Just keep walking the matrix, when hitting the four borders (top, bottom, left or right),
+         * just directions and keep walking.
+         */
+        public int[] findDiagonalOrder(int[][] matrix) {
 
-        if (matrix == null || matrix.length == 0) {
-            return new int[0];
-        }
-        int m = matrix.length;
-        int n = matrix[0].length;
-        int[] result = new int[m * n];
-        int d = 1;
-        int i = 0;
-        int j = 0;
-        for (int k = 0; k < m * n; ) {
-            result[k++] = matrix[i][j];
-            i -= d;
-            j += d;
-
-            if (i >= m) {
-                i = m - 1;
-                j += 2;
-                d = -d;
-            }
-            if (j >= n) {
-                j = n - 1;
-                i += 2;
-                d = -d;
+            if (matrix == null || matrix.length == 0) {
+                return new int[0];
             }
-            if (i < 0) {
-                i = 0;
-                d = -d;
-            }
-            if (j < 0) {
-                j = 0;
-                d = -d;
+            int m = matrix.length;
+            int n = matrix[0].length;
+            int[] result = new int[m * n];
+            int d = 1;
+            int i = 0;
+            int j = 0;
+            for (int k = 0; k < m * n; ) {
+                result[k++] = matrix[i][j];
+                i -= d;
+                j += d;
+
+                if (i >= m) {
+                    i = m - 1;
+                    j += 2;
+                    d = -d;
+                }
+                if (j >= n) {
+                    j = n - 1;
+                    i += 2;
+                    d = -d;
+                }
+                if (i < 0) {
+                    i = 0;
+                    d = -d;
+                }
+                if (j < 0) {
+                    j = 0;
+                    d = -d;
+                }
             }
+            return result;
         }
-        return result;
     }
-}
 
 }