From 6ff87322e63e167cf62dc1bb94eb426c27a52649 Mon Sep 17 00:00:00 2001
From: Libin Yang <szuyanglb@outlook.com>
Date: Sat, 23 Feb 2019 21:06:32 +0800
Subject: [PATCH] docs(DP): update EggDropping.java

---
 Dynamic Programming/EggDropping.java | 98 +++++++++++++---------------
 1 file changed, 46 insertions(+), 52 deletions(-)

diff --git a/Dynamic Programming/EggDropping.java b/Dynamic Programming/EggDropping.java
index 382d7ea8..0549f664 100644
--- a/Dynamic Programming/EggDropping.java	
+++ b/Dynamic Programming/EggDropping.java	
@@ -1,53 +1,47 @@
-//Dynamic Programming solution for the Egg Dropping Puzzle
-public class EggDropping
-{
-	
-	// min trials with n eggs and m floors 
-	
-	private static int minTrials(int n, int m)
-	{
-	    
-		int eggFloor[][] = new int[n+1][m+1];
-		int result, x;
-		
-		for (int i = 1; i <= n; i++)
-		{
-		    eggFloor[i][0] = 0;   // Zero trial for zero floor.
-		    eggFloor[i][1] = 1;   // One trial for one floor 
-		}
-		
-	        // j trials for only 1 egg
-		
-                for (int j = 1; j <= m; j++)
-			 eggFloor[1][j] = j;
-		
-		// Using bottom-up approach in DP
-		
-   		 for (int i = 2; i <= n; i++)
-		  {
-			for (int j = 2; j <= m; j++)
-			{
-				eggFloor[i][j] = Integer.MAX_VALUE;
-				for (x = 1; x <= j; x++)
-				{
-					result = 1 + Math.max(eggFloor[i-1][x-1], eggFloor[i][j-x]);
-          
-					//choose min of all values for particular x
-        				 if (result < eggFloor[i][j])
-					     eggFloor[i][j] = result;
-				}
-			}
-		}
-		  
-		return eggFloor[n][m];
-	}
-		
-	//testing program
-	public static void main(String args[])
-	{
-	      int n = 2, m = 4;
-              //result outputs min no. of trials in worst case for n eggs and m floors
-              int result = minTrials(n, m);
-	      System.out.println(result); 
-	}
+/**
+ * Dynamic Programming solution for the Egg Dropping Puzzle
+ */
+public class EggDropping {
+
+    // min trials with n eggs and m floors 
+
+    private static int minTrials(int n, int m) {
+
+        int[][] eggFloor = new int[n + 1][m + 1];
+        int result, x;
+
+        for (int i = 1; i <= n; i++) {
+            eggFloor[i][0] = 0;   // Zero trial for zero floor.
+            eggFloor[i][1] = 1;   // One trial for one floor 
+        }
+
+        // j trials for only 1 egg
+
+        for (int j = 1; j <= m; j++)
+            eggFloor[1][j] = j;
+
+        // Using bottom-up approach in DP
+
+        for (int i = 2; i <= n; i++) {
+            for (int j = 2; j <= m; j++) {
+                eggFloor[i][j] = Integer.MAX_VALUE;
+                for (x = 1; x <= j; x++) {
+                    result = 1 + Math.max(eggFloor[i - 1][x - 1], eggFloor[i][j - x]);
+
+                    // choose min of all values for particular x
+                    if (result < eggFloor[i][j])
+                        eggFloor[i][j] = result;
+                }
+            }
+        }
+
+        return eggFloor[n][m];
+    }
+    
+    public static void main(String args[]) {
+        int n = 2, m = 4;
+        // result outputs min no. of trials in worst case for n eggs and m floors
+        int result = minTrials(n, m);
+        System.out.println(result);
+    }
 }