refactor 390

Author: fishercoder1534

src/main/java/com/fishercoder/solutions/_390.java

@@ -1,9 +1,9 @@
 package com.fishercoder.solutions;
 
-import java.util.ArrayList;
-import java.util.List;
-
-/**There is a list of sorted integers from 1 to n.
+/**
+ * 390. Elimination Game
+ *
+ * There is a list of sorted integers from 1 to n.
  * Starting from left to right,
  * remove the first number and every other number afterward until you reach the end of the list.
  * Repeat the previous step again, but this time from right to left,
@@ -25,68 +25,29 @@
 
  */
 public class _390 {
-    
-    //then I turned to Discuss and found this post: https://discuss.leetcode.com/topic/55870/share-my-solutions-for-contest-2
-    //instead of literally removing half of the elements in each scan, this solution is just moving the pointer to point to next start position
-    //So brilliant!
-    public int lastRemaining(int n) {
-        int remaining = n;
-        int start = 1;
-        int step = 2;
-        boolean forward = true;
-        while (remaining > 1) {
-            remaining /= 2;
-            if (forward) {
-                start = start + step * remaining - step / 2;
-            } else {
-                start = start - step * remaining + step / 2;
-            }
-            step *= 2;
-            forward = !forward;
-        }
-        return start;
-    }
 
-    //I tried brute force, all producing the correct output, but got TLE by OJ.
-    public static int lastRemaining_brute_force_TLE(int n) {
-        List<Integer> list = new ArrayList();
-        for (int i = 0; i < n; i++) {
-            list.add(i + 1);
-        }
-        boolean forward = true;
-        while (list.size() > 1) {
-            int size = list.size() / 2;
-            if (list.size() == 1) {
-                return list.get(0);
-            }
-            if (forward) {
-                if (list.size() == 1) {
-                    return list.get(0);
+    public static class Solution1 {
+        /**
+         * credit: https://discuss.leetcode.com/topic/55870/share-my-solutions-for-contest-2 instead of
+         * literally removing half of the elements in each scan, this solution is just moving the
+         * pointer to point to next start position So brilliant!
+         */
+        public int lastRemaining(int n) {
+            int remaining = n;
+            int start = 1;
+            int step = 2;
+            boolean forward = true;
+            while (remaining > 1) {
+                remaining /= 2;
+                if (forward) {
+                    start = start + step * remaining - step / 2;
+                } else {
+                    start = start - step * remaining + step / 2;
                 }
-                for (int i = 0; i <= size && i < list.size(); i++) {
-                    list.remove(i);
-                }
-                forward = false;
-            } else {
-                if (list.size() == 1) {
-                    return list.get(0);
-                }
-                for (int i = list.size() - 1, count = 0; i >= 0 && count <= size; count++) {
-                    list.remove(i);
-                    i -= 2;
-                }
-                forward = true;
+                step *= 2;
+                forward = !forward;
             }
+            return start;
         }
-        return list.get(0);
     }
-
-    public static void main(String... strings) {
-        System.out.println(lastRemaining_brute_force_TLE(5204));
-        System.out.println(lastRemaining_brute_force_TLE(5058));
-//        System.out.println(lastRemaining(10));
-//        System.out.println(lastRemaining(9));
-//        System.out.println(lastRemaining(3));
-    }
-
 }