refactor 573

fishercoder1534 · fishercoder1534 · commit 5ff0d3d49e88 · 2019-07-11T07:07:09.000-07:00
diff --git a/src/main/java/com/fishercoder/solutions/_573.java b/src/main/java/com/fishercoder/solutions/_573.java
@@ -33,33 +33,37 @@
  */
 public class _573 {
 
-    /**reference: https://leetcode.com/articles/squirrel-simulation
-     *
-     * 1. The order in which to pick the nuts does not matter except the first one
-     * because for all the other nuts, the squirrel needs to travel back and forth.
-     *
-     * 2. For the first nut to be picked, there's some distance we might be able to save, what is this distance?
-     * It is the difference between the squirrel's original starting point to the first nut and that the distance from this
-     * first nut to the tree.
-     * This is because, only for the first nut, the squirrel does NOT need to travel back and forth, it only needs to travel from
-     * its starting position to the nut position and then return to the tree.
-     *
-     * 3. For the rest of all other nuts, the squirrel always needs to go back and forth.
-     *
-     * 4. So how can we find the first nut to go to so that we could have the maximum saved distance?
-     * We iterate through all of the nuts and keep updating the savedDist as below:
-     * */
-    public int minDistance(int height, int width, int[] tree, int[] squirrel, int[][] nuts) {
-        int totalDist = 0;
-        int savedDist = Integer.MIN_VALUE;
-        for (int[] nut : nuts) {
-            totalDist += (getDist(nut, tree) * 2);//it needs to travel back and forth, so times two
-            savedDist = Math.max(savedDist, getDist(nut, tree) - getDist(nut, squirrel));
+    public static class Solution1 {
+
+        /**
+         * reference: https://leetcode.com/articles/squirrel-simulation
+         *
+         * 1. The order in which to pick the nuts does not matter except the first one
+         * because for all the other nuts, the squirrel needs to travel back and forth.
+         *
+         * 2. For the first nut to be picked, there's some distance we might be able to save, what is this distance?
+         * It is the difference between the squirrel's original starting point to the first nut and that the distance from this
+         * first nut to the tree.
+         * This is because, only for the first nut, the squirrel does NOT need to travel back and forth, it only needs to travel from
+         * its starting position to the nut position and then return to the tree.
+         *
+         * 3. For the rest of all other nuts, the squirrel always needs to go back and forth.
+         *
+         * 4. So how can we find the first nut to go to so that we could have the maximum saved distance?
+         * We iterate through all of the nuts and keep updating the savedDist as below:
+         */
+        public int minDistance(int height, int width, int[] tree, int[] squirrel, int[][] nuts) {
+            int totalDist = 0;
+            int savedDist = Integer.MIN_VALUE;
+            for (int[] nut : nuts) {
+                totalDist += (getDist(nut, tree) * 2);//it needs to travel back and forth, so times two
+                savedDist = Math.max(savedDist, getDist(nut, tree) - getDist(nut, squirrel));
+            }
+            return totalDist - savedDist;
         }
-        return totalDist - savedDist;
-    }
 
-    private int getDist(int[] pointA, int[] pointB) {
-        return Math.abs(pointA[0] - pointB[0]) + Math.abs(pointA[1] - pointB[1]);
+        private int getDist(int[] pointA, int[] pointB) {
+            return Math.abs(pointA[0] - pointB[0]) + Math.abs(pointA[1] - pointB[1]);
+        }
     }
 }
