refactor 533

Author: fishercoder1534

src/main/java/com/fishercoder/solutions/_533.java

@@ -38,53 +38,55 @@
  The range of width and height of the input 2D array is [1,200].
  */
 public class _533 {
-    /**Credit: https://discuss.leetcode.com/topic/81686/verbose-java-o-m-n-solution-hashmap/5
-     *
-     * This program is very well designed to do things:
-     * 1. it scans the entire matrix once, but does two things in this scan:
-     * first it records an array of cols that keeps count of how many 'B' are there in each column;
-     * second, at the end of traversing each column, it puts this entire row as the key into a HashMap
-     * when there N number of 'B's in this row.
-     *
-     * 2. then we could go through the HashMap keyset:
-     * if one row has N number of 'B's, we go through this row's each column to see if any element in this row is 'B' and also that element's column has N 'B's*/
-    public int findBlackPixel(char[][] picture, int N) {
-        if (picture == null || picture.length == 0 || picture[0].length == 0) {
-            return 0;
-        }
-        int m = picture.length;
-        int n = picture[0].length;
-        int[] cols = new int[n];
-        Map<String, Integer> map = new HashMap<>();
-        StringBuilder stringBuilder = new StringBuilder();
-        for (int i = 0; i < m; i++) {
-            int count  = 0;
-            for (int j = 0; j < n; j++) {
-                if (picture[i][j] == 'B') {
-                    count++;
-                    cols[j]++;
-                }
-                stringBuilder.append(picture[i][j]);
+    public static class Solution1 {
+        /**
+         * Credit: https://discuss.leetcode.com/topic/81686/verbose-java-o-m-n-solution-hashmap/5
+         * This program is very well designed to do things:
+         * 1. it scans the entire matrix once, but does two things in this scan:
+         * first it records an array of cols that keeps count of how many 'B' are there in each column;
+         * second, at the end of traversing each column, it puts this entire row as the key into a HashMap
+         * when there N number of 'B's in this row.
+         * 2. then we could go through the HashMap keyset:
+         * if one row has N number of 'B's, we go through this row's each column to see if any element in this row is 'B' and also that element's column has N 'B's
+         */
+        public int findBlackPixel(char[][] picture, int N) {
+            if (picture == null || picture.length == 0 || picture[0].length == 0) {
+                return 0;
             }
-            if (count == N) {
-                /**we use this entire row string as key for the map*/
-                map.put(stringBuilder.toString(), map.getOrDefault(stringBuilder.toString(), 0) + 1);
+            int m = picture.length;
+            int n = picture[0].length;
+            int[] cols = new int[n];
+            Map<String, Integer> map = new HashMap<>();
+            StringBuilder stringBuilder = new StringBuilder();
+            for (int i = 0; i < m; i++) {
+                int count = 0;
+                for (int j = 0; j < n; j++) {
+                    if (picture[i][j] == 'B') {
+                        count++;
+                        cols[j]++;
+                    }
+                    stringBuilder.append(picture[i][j]);
+                }
+                if (count == N) {
+                    /**we use this entire row string as key for the map*/
+                    map.put(stringBuilder.toString(), map.getOrDefault(stringBuilder.toString(), 0) + 1);
+                }
+                stringBuilder.setLength(0);
             }
-            stringBuilder.setLength(0);
-        }
 
-        int answer = 0;
-        for (String key : map.keySet()) {
-            if (map.get(key) != N) {
-                continue;
-            }
-            for (int i = 0; i < n; i++) {
-                if (key.charAt(i) == 'B' && cols[i] == N) {
-                    answer += N;
+            int answer = 0;
+            for (String key : map.keySet()) {
+                if (map.get(key) != N) {
+                    continue;
+                }
+                for (int i = 0; i < n; i++) {
+                    if (key.charAt(i) == 'B' && cols[i] == N) {
+                        answer += N;
+                    }
                 }
             }
+            return answer;
         }
-        return answer;
     }
 
 }

src/test/java/com/fishercoder/_533Test.java

@@ -6,16 +6,13 @@
 
 import static org.junit.Assert.assertEquals;
 
-/**
- * Created by fishercoder on 5/25/17.
- */
 public class _533Test {
-    private static _533 test;
+    private static _533.Solution1 solution1;
     private static char[][] picture;
 
     @BeforeClass
     public static void setup() {
-        test = new _533();
+        solution1 = new _533.Solution1();
     }
 
     @Test
@@ -25,12 +22,12 @@ public void test1() {
                 {'W', 'B', 'W', 'B', 'B', 'W'},
                 {'W', 'B', 'W', 'B', 'B', 'W'},
                 {'W', 'W', 'B', 'W', 'B', 'W'}};
-        assertEquals(6, test.findBlackPixel(picture, 3));
+        assertEquals(6, solution1.findBlackPixel(picture, 3));
     }
 
     @Test
     public void test2() {
         picture = new char[][]{{'B'}};
-        assertEquals(1, test.findBlackPixel(picture, 1));
+        assertEquals(1, solution1.findBlackPixel(picture, 1));
     }
 }