regular expression

Author: awangdev

Java/Delete Digits.java

@@ -1,5 +1,6 @@
 M
-Tags: Greedy, Priority Queue
+1534317486
+tags: Greedy, Priority Queue
 
 #### Priority Queue
 - TODO: parse into node(index, digitValue)

Java/Flatten 2D Vector.java

@@ -1,8 +1,11 @@
 M
+1534317715
 tags: Design
 
 Implement an iterator to flatten a 2d vector.
 
+Just move pointers carefully with next(), hashNext()
+
 #### Basic Implementation using x, y corrdinate
 - 就是把2D list里面的element全部遍历一遍。
 - 跟一个nxn的matrix遍历，是没区别的拉; 所有来个x,y，把2d list跑一变。

Java/Longest Substring with At Most K Distinct Characters.java

@@ -1,6 +1,6 @@
-M
+H
 1520228394
-tags: Hash Table, String
+tags: Hash Table, String, Sliding Window
 
 大清洗 O(nk)   
 map.size一旦>k，要把longest string最开头（marked by pointer:start）的那个char抹掉    
@@ -36,27 +36,23 @@ public int lengthOfLongestSubstringKDistinct(String s, int k) {
             return 0;
         }
         int n = s.length();
-        int start = 0;
-        int end = 0;
-        int rst = Integer.MIN_VALUE;
+        int start = 0, end = 0, rst = Integer.MIN_VALUE;
         HashMap<Character, Integer> map = new HashMap<>();
-        while (start < n) {
+        while (start < n) { // left window
             while (end < n) {
                 char c = s.charAt(end);
                 if (map.containsKey(c)) {
                     map.put(c, map.get(c) + 1);
                 } else {
-                    if (map.size() == k) {
-                        break;
-                    }
+                    if (map.size() == k) break; // meet window size
                     map.put(c, 1);
                 }                
                 end++;
             }
-            // Calculate length when map.size() == n or end == n
+            // Calculate length when map.size() == k or end == n
             rst = Math.max(rst, end - start);
 
-            // move start forward
+            // move start forward and clean up map
             char c = s.charAt(start);
             int count = map.get(c);
             if (count == 1) {
@@ -71,7 +67,6 @@ public int lengthOfLongestSubstringKDistinct(String s, int k) {
     }
 }
 
-
 /*
 Thoughts:
 HashMap size cannot go over k.

Java/Regular Expression Matching.java

@@ -1,6 +1,12 @@
-R
-1519287323
-tags: String, DP, Backtracking
+H
+1534348524
+tags: String, DP, Sequence DP, Double Sequence DP, Backtracking
+
+跟WildCard Matching 一样, 分清楚情况讨论 string p last char is '*' 还有并不是 '*'
+
+这里的区别是, '*' 需要有一个preceding element, 那么:
+- repeat 0 times
+- repeat 1 times: need s[i-1] match with prior char p[i-2]
 
 ```
 /*
@@ -42,14 +48,8 @@ bool isMatch(const char *s, const char *p)
 */
 class Solution {
     public boolean isMatch(String s, String p) {
-        if (s == null || p == null) {
-            return false;
-        }
-        if (s.length() != p.length() && p.indexOf('.') < 0 && p.indexOf('*') < 0) {
-            return false;
-        }
-        int m = s.length();
-        int n = p.length();
+        if (s == null || p == null) return false;
+        int m = s.length(), n = p.length();
         boolean[][] dp = new boolean[m + 1][n + 1];
         char[] ss = s.toCharArray();
         char[] pp = p.toCharArray();
@@ -60,7 +60,7 @@ public boolean isMatch(String s, String p) {
                     dp[i][j] = true;
                     continue;
                 }
-                if (i >= 1 && j == 0) {
+                if (j == 0) { // When p is empty but s is not empty, should not match
                     dp[i][j] = false;
                     continue;
                 }
@@ -72,10 +72,11 @@ public boolean isMatch(String s, String p) {
                         dp[i][j] = dp[i - 1][j - 1];
                     }
                 } else { // tail = '*'. ex: a*
-                    if (j >= 2 ) { // ignore a*
-                        dp[i][j] = dp[i][j - 2];
+                    if (j >= 2 ) { // ignore a*, repeat 0 times
+                        dp[i][j] |= dp[i][j - 2];
                     }
-                    if (j >= 2 && i >= 1 && (pp[j - 2] == '.' || pp[j - 2] == ss[i - 1])) {
+                    // repeat the char befeore * for 1 time, so ss[i-1] should match pp[j-2] or pp[j-2] == '.'
+                    if (j >= 2 && i >= 1 && (ss[i - 1] == pp[j - 2] || pp[j - 2] == '.')) { 
                         dp[i][j] |= dp[i - 1][j];
                     }
                 }

Java/The Spiral Matrix II.java

@@ -0,0 +1,51 @@
+M
+1534346957
+tags: Array
+
+#### Move forward till end
+- Similar concept as `The Maze`: keep walking until hit wall, turn back
+- fix direction `dx[direction % 4]`
+
+```
+/*
+Given a positive integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.
+
+Example:
+
+Input: 3
+Output:
+[
+ [ 1, 2, 3 ],
+ [ 8, 9, 4 ],
+ [ 7, 6, 5 ]
+]
+*/
+
+/*
+Use while loop, reach end and stop
+*/
+class Solution {
+    int[] dx = {0, 1, 0, -1};
+    int[] dy = {1, 0, -1, 0};
+    public int[][] generateMatrix(int n) {
+        int[][] grid = new int[n][n];
+        int step = 1, i = 0, j = 0, direction = 0;
+        grid[i][j] = step++;
+        while (step <= n * n) {
+            int x = dx[direction % 4];
+            int y = dy[direction % 4];
+            i += x;
+            j += y;
+            while (i >= 0 && i < n && j >= 0 && j < n && grid[i][j] == 0) {
+                grid[i][j] = step++;
+                i += x;
+                j += y;
+            }
+            i -= x;
+            j -= y;
+            direction++;
+        }
+        return grid;
+    }
+}
+```

Java/Wildcard Matching.java

@@ -1,12 +1,20 @@
 H
-1519368767
-tags: String, DP, Backtracking, Greedy
+1534348528
+tags: String, DP, Sequence DP, Double Sequence DP, Backtracking, Greedy
 
 Double sequence DP. 与regular expression 很像.
 
-注意1: 分析字符 ?, * 所代表的真正意义, 然后写出表达式.
-注意2: 搞清楚initialization 的时候 dp[i][0] 应该always false.当p为empty string, 无论如何都match不了 (除非s="" as well)
-    同时 dp[0][j]不一定是false. 比如s="",p="*" 就是一个matching.
+#### Double Sequence DP
+- 分析字符 ?, * 所代表的真正意义, 然后写出表达式.
+- 搞清楚initialization 的时候 dp[i][0] 应该always false. 当p为empty string, 无论如何都match不了 (除非s="" as well)
+- 同时 dp[0][j]不一定是false. 比如s="",p="*" 就是一个matching.
+- A. p[j] != '*'
+    1. last index match => dp[i - 1][j - 1]
+    2. last index == ?  => dp[i - 1][j - 1]
+- B. p[j] == "*"
+    1. * is empty => dp[i][j - 1]
+    2. * match 1 or more chars => dp[i - 1][j]
+
 
 ```
 /*
@@ -51,17 +59,12 @@ Time,Space O(MN)
 */
 class Solution {
     public boolean isMatch(String s, String p) {
-        if (s == null || p == null) {
-            return false;
-        }
-        int m = s.length();
-        int n = p.length();
+        if (s == null || p == null) return false;
+        int m = s.length(), n = p.length();
         boolean[][] dp = new boolean[m + 1][n + 1];
         char[] ss = s.toCharArray();
         char[] pp = p.toCharArray();
 
-        // dp[0][j] = false; dp[i][0] = false; 
-        
         for (int i = 0; i <= m; i++) {
             for (int j = 0; j <= n; j++) {
                 if (i == 0 && j == 0) {
@@ -75,7 +78,7 @@ public boolean isMatch(String s, String p) {
                 dp[i][j] = false;
                 if (pp[j - 1] != '*') {
                     if (i >= 1 && (ss[i - 1] == pp[j - 1] || pp[j - 1] == '?')) {
-                        dp[i][j] = dp[i - 1][j - 1];
+                        dp[i][j] |= dp[i - 1][j - 1];
                     }
                 } else {
                     dp[i][j] |= dp[i][j - 1];// '*' -> empty