refactor 85

Author: fishercoder1534

src/main/java/com/fishercoder/solutions/_85.java

@@ -2,74 +2,60 @@
 
 import java.util.Arrays;
 
-/**
- * 85. Maximal Rectangle
- *
- * Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing only 1's and return its area.
-
- For example, given the following matrix:
-
- 1 0 1 0 0
- 1 0 1 1 1
- 1 1 1 1 1
- 1 0 0 1 0
- Return 6.
-
- */
 public class _85 {
-  public static class Solution1 {
-    public int maximalRectangle(char[][] matrix) {
-      if (matrix.length == 0) {
-        return 0;
-      }
-      int m = matrix.length;
-      int n = matrix[0].length;
-      int[] left = new int[n];
-      int[] right = new int[n];
-      int[] height = new int[n];
-      Arrays.fill(left, 0);
-      Arrays.fill(right, n);
-      Arrays.fill(height, 0);
-      int maxA = 0;
-      for (int i = 0; i < m; i++) {
-        int currLeft = 0;
-        int currRight = n;
-
-        //compute height, this can be achieved from either side
-        for (int j = 0; j < n; j++) {
-          if (matrix[i][j] == '1') {
-            height[j]++;
-          } else {
-            height[j] = 0;
-          }
-        }
-
-        //compute left, from left to right
-        for (int j = 0; j < n; j++) {
-          if (matrix[i][j] == '1') {
-            left[j] = Math.max(left[j], currLeft);
-          } else {
-            left[j] = 0;
-            currLeft = j + 1;
-          }
-        }
-
-        //compute right, from right to left
-        for (int j = n - 1; j >= 0; j--) {
-          if (matrix[i][j] == '1') {
-            right[j] = Math.min(right[j], currRight);
-          } else {
-            right[j] = n;
-            currRight = j;
-          }
-        }
-
-        //compute rectangle area, this can be achieved from either side
-        for (int j = 0; j < n; j++) {
-          maxA = Math.max(maxA, (right[j] - left[j]) * height[j]);
+    public static class Solution1 {
+        public int maximalRectangle(char[][] matrix) {
+            if (matrix.length == 0) {
+                return 0;
+            }
+            int m = matrix.length;
+            int n = matrix[0].length;
+            int[] left = new int[n];
+            int[] right = new int[n];
+            int[] height = new int[n];
+            Arrays.fill(left, 0);
+            Arrays.fill(right, n);
+            Arrays.fill(height, 0);
+            int maxA = 0;
+            for (int i = 0; i < m; i++) {
+                int currLeft = 0;
+                int currRight = n;
+
+                //compute height, this can be achieved from either side
+                for (int j = 0; j < n; j++) {
+                    if (matrix[i][j] == '1') {
+                        height[j]++;
+                    } else {
+                        height[j] = 0;
+                    }
+                }
+
+                //compute left, from left to right
+                for (int j = 0; j < n; j++) {
+                    if (matrix[i][j] == '1') {
+                        left[j] = Math.max(left[j], currLeft);
+                    } else {
+                        left[j] = 0;
+                        currLeft = j + 1;
+                    }
+                }
+
+                //compute right, from right to left
+                for (int j = n - 1; j >= 0; j--) {
+                    if (matrix[i][j] == '1') {
+                        right[j] = Math.min(right[j], currRight);
+                    } else {
+                        right[j] = n;
+                        currRight = j;
+                    }
+                }
+
+                //compute rectangle area, this can be achieved from either side
+                for (int j = 0; j < n; j++) {
+                    maxA = Math.max(maxA, (right[j] - left[j]) * height[j]);
+                }
+            }
+            return maxA;
         }
-      }
-      return maxA;
     }
-  }
 }