refactor 85

fishercoder1534 · fishercoder1534 · commit b038d8bb6a32 · 2018-02-22T09:21:25.000-08:00
diff --git a/src/main/java/com/fishercoder/solutions/_85.java b/src/main/java/com/fishercoder/solutions/_85.java
@@ -3,6 +3,8 @@
 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:
@@ -12,59 +14,62 @@
  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;
+      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 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 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 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]);
-            }
+        //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;
     }
+  }
 }
