refactor 378

Author: fishercoder1534

README.md

@@ -249,6 +249,7 @@ Your ideas/fixes/algorithms are more than welcome!
 |381|[Insert Delete GetRandom O(1) - Duplicates allowed](https://leetcode.com/problems/insert-delete-getrandom-o1-duplicates-allowed/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_381.java)| | | Hard| 
 |380|[Insert Delete GetRandom O(1)](https://leetcode.com/problems/insert-delete-getrandom-o1/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_380.java)| O(n) | O(1)| Medium| Design, HashMap
 |379|[Design Phone Directory](https://leetcode.com/problems/design-phone-directory/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_379.java)| O(1)|O(n) | Medium| 
+|378|[Kth Smallest Element in a Sorted Matrix](https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_378.java)| O(logm*n) | O(1)| Medium| Binary Search
 |377|[Combination Sum IV](https://leetcode.com/problems/combination-sum-iv/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_377.java)| O(?)|O(?) | Medium|
 |376|[Wiggle Subsequence](https://leetcode.com/problems/wiggle-subsequence/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_376.java)| O(n)|O(1) | Medium| DP, Greedy
 |375|[Guess Number Higher or Lower II](https://leetcode.com/problems/guess-number-higher-or-lower-ii/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_375.java)| O(n^2)|O(n^2) | Medium| DP

src/main/java/com/fishercoder/solutions/_378.java

@@ -3,9 +3,11 @@
 import java.util.ArrayList;
 import java.util.Collections;
 import java.util.List;
-/**378. Kth Smallest Element in a Sorted Matrix
- *
-Given a n x n matrix where each of the rows and columns are sorted in ascending order, find the kth smallest element in the matrix.
+
+/**
+ * 378. Kth Smallest Element in a Sorted Matrix
+ * Given a n x n matrix where each of the rows and columns are
+ * sorted in ascending order, find the kth smallest element in the matrix.
 
 Note that it is the kth smallest element in the sorted order, not the kth distinct element.
 
@@ -20,48 +22,56 @@
 
 return 13.
 
- Note:
-You may assume k is always valid, 1 ≤ k ≤ n2.*/
+Note:
+You may assume k is always valid, 1 ≤ k ≤ n2.
+ */
+
 public class _378 {
-    //brute force made it AC'ed, extreme test case needed for OJ
-    public int kthSmallest(int[][] matrix, int k) {
-        List<Integer> list = new ArrayList<Integer>();
-        int n = matrix.length;
-        for (int i = 0; i < n; i++) {
-            for (int j = 0; j < n; j++) {
-                list.add(matrix[i][j]);
+    public static class Solution1 {
+        /**
+         * brute force made it AC'ed, extreme test case needed for OJ
+         */
+        public int kthSmallest(int[][] matrix, int k) {
+            List<Integer> list = new ArrayList();
+            int n = matrix.length;
+            for (int i = 0; i < n; i++) {
+                for (int j = 0; j < n; j++) {
+                    list.add(matrix[i][j]);
+                }
             }
+            Collections.sort(list);
+            return list.get(k - 1);
         }
-        Collections.sort(list);
-        return list.get(k - 1);
     }
 
-    //TODO: use heap and binary search to do it.
-
-    //Binary Search : The idea is to pick a mid number than compare it with the elements in each row, we start form
-    // end of row util we find the element is less than the mid, the left side element is all less than mid; keep tracking elements
-    // that less than mid and compare with k, then update the k.
-    public int kthSmallestBS(int[][] matrix, int k) {
-        int row = matrix.length - 1;
-        int col = matrix[0].length - 1;
-        int lo = matrix[0][0];
-        int hi = matrix[row][col];
-        while (lo < hi) {
-            int mid = lo + (hi - lo) / 2;
-            int count = 0;
-            int j = col;
-            for (int i = 0; i <= row; i++) {
-                while (j >= 0 && matrix[i][j] > mid) {
-                    j--;
+    public static class Solution2 {
+        /**
+         * Binary Search : The idea is to pick a mid number, then compare it with the elements in each row, we start form
+         * end of row util we find the element is less than the mid, the left side element is all less than mid; keep tracking elements
+         * that less than mid and compare with k, then update the k.
+         */
+        public int kthSmallestBS(int[][] matrix, int k) {
+            int row = matrix.length - 1;
+            int col = matrix[0].length - 1;
+            int lo = matrix[0][0];
+            int hi = matrix[row][col];
+            while (lo < hi) {
+                int mid = lo + (hi - lo) / 2;
+                int count = 0;
+                int j = col;
+                for (int i = 0; i <= row; i++) {
+                    while (j >= 0 && matrix[i][j] > mid) {
+                        j--;
+                    }
+                    count += (j + 1);
+                }
+                if (count < k) {
+                    lo = mid + 1;
+                } else {
+                    hi = mid;
                 }
-                count += (j + 1);
-            }
-            if (count < k) {
-                lo = mid + 1;
-            } else {
-                hi = mid;
             }
+            return lo;
         }
-        return lo;
     }
 }