refactor 84

Author: fishercoder1534

src/main/java/com/fishercoder/solutions/_84.java

@@ -2,43 +2,30 @@
 
 import java.util.Stack;
 
-/**
- * 84. Largest Rectangle in Histogram
- *
- * Given n non-negative integers representing the histogram's bar height where
- * the width of each bar is 1, find the area of largest rectangle in the histogram.
-
- Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3].
- The largest rectangle is shown in the shaded area, which has area = 10 unit.
-
- For example,
- Given heights = [2,1,5,6,2,3],
- return 10.
- */
 public class _84 {
 
-  public static class Solution1 {
+    public static class Solution1 {
 
-    /**
-     * credit: https://leetcode.com/articles/largest-rectangle-histogram/#approach-5-using-stack-accepted
-     * and https://discuss.leetcode.com/topic/7599/o-n-stack-based-java-solution
-     */
-    public int largestRectangleArea(int[] heights) {
-      int len = heights.length;
-      Stack<Integer> s = new Stack<>();
-      int maxArea = 0;
-      for (int i = 0; i <= len; i++) {
-        int h = (i == len ? 0 : heights[i]);
-        if (s.isEmpty() || h >= heights[s.peek()]) {
-          s.push(i);
-        } else {
-          int tp = s.pop();
-          maxArea = Math.max(maxArea, heights[tp] * (s.isEmpty() ? i : i - 1 - s.peek()));
-          i--;
+        /**
+         * credit: https://leetcode.com/articles/largest-rectangle-histogram/#approach-5-using-stack-accepted
+         * and https://discuss.leetcode.com/topic/7599/o-n-stack-based-java-solution
+         */
+        public int largestRectangleArea(int[] heights) {
+            int len = heights.length;
+            Stack<Integer> s = new Stack<>();
+            int maxArea = 0;
+            for (int i = 0; i <= len; i++) {
+                int h = (i == len ? 0 : heights[i]);
+                if (s.isEmpty() || h >= heights[s.peek()]) {
+                    s.push(i);
+                } else {
+                    int tp = s.pop();
+                    maxArea = Math.max(maxArea, heights[tp] * (s.isEmpty() ? i : i - 1 - s.peek()));
+                    i--;
+                }
+            }
+            return maxArea;
         }
-      }
-      return maxArea;
     }
-  }
 
 }