Merge branch 'master' of https://github.com/azl397985856/leetcode

Author: luzhipeng

daily/2019-06-09.md

@@ -0,0 +1,134 @@
+## 每日一题 - Regular Expression Matching
+
+### 信息卡片
+
+- 时间：2019-06-09
+- 题目链接：https://leetcode.com/problems/regular-expression-matching/
+- tag：`String` `Dynamic Programming` `Backtracking`
+
+### 题目描述
+
+```
+Given an input string (s) and a pattern (p), implement regular expression matching with support for '.' and '*'.
+
+
+'.' Matches any single character.
+'*' Matches zero or more of the preceding element.
+
+
+The matching should cover the entire input string (not partial).
+
+Note:
+
+
+s could be empty and contains only lowercase letters a-z.
+p could be empty and contains only lowercase letters a-z, and characters
+like . or *.
+
+
+Example 1:
+
+
+Input:
+s = "aa"
+p = "a"
+Output: false
+Explanation: "a" does not match the entire string "aa".
+
+
+Example 2:
+
+
+Input:
+s = "aa"
+p = "a*"
+Output: true
+Explanation: '*' means zero or more of the precedeng element, 'a'.
+Therefore, by repeating 'a' once, it becomes "aa".
+
+
+Example 3:
+
+
+Input:
+s = "ab"
+p = ".*"
+Output: true
+Explanation: ".*" means "zero or more (*) of any character (.)".
+
+
+Example 4:
+
+
+Input:
+s = "aab"
+p = "c*a*b"
+Output: true
+Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore
+it matches "aab".
+
+
+Example 5:
+
+
+Input:
+s = "mississippi"
+p = "mis*is*p*."
+Output: false
+
+```
+
+本题要求判断给出的字符串和对应的正则是否匹配，匹配返回true，否则返回false。
+
+本题光从解题来看，可以使用api作弊，比如用Java字符串的matches方法就可以一步过这道题，本题老老实实做的话，就需要用到动态规划。基本思路是考察s和p任意从头到两个字符之间的匹配程度，有点像编辑距离那道题。
+
+### 参考答案
+
+```java
+class Solution {
+    public boolean isMatch(String s, String p) {
+      if (s == null || p == null) return false;
+ 
+      // dp[i][j]代表s的前i位字符和p的前j位字符的匹配程度
+      // dp[1][2]代表s的第一个字符和p的前两个字符的匹配
+      // 如果s的第i个字符和p的第j个相等或者j为.那么di[i][j]的匹配程度取决于dp[i - 1][j - 1]
+      // 如果p的第j个字符为*，那么就要考虑*匹配0个，1个，n个s的第i个字符的情况，以及根本匹配不了的情况
+      // 根本匹配不了的意思是指p的j - 1个字符和s的第i个字符不相同，此时的匹配情况是dp[i][j] = dp[i][j - 2]
+      // 然后是p的第j - 1个字符是.或者和s的i个字符相同的情况，此时可以匹配0 1 个n个s的第i个字符
+      // 0个: dp[i][j] = dp[i][j - 2]
+      // 1个: dp[i][j] = dp[i - 1][j - 1]
+      // n个: dp[i][j] = dp[i - 1][j]
+      // n个的最不好理解，可以按照下面的想
+      // 我能匹配你n个，我肯定匹配了你前面的n-1个，也就是dp[i - 1][j]
+      // 最后的结果是dp[s.length()][p.length()]
+      boolean[][] dp = new boolean[s.length() + 1][p.length() + 1];
+
+      dp[0][0] = true;
+
+      for (int j = 2; j <= p.length(); j++) {
+        if (p.charAt(j - 1) ==  '*' && dp[0][j - 2]) dp[0][j] = true;
+      }
+
+      for (int i = 1; i <= s.length(); i++) {
+        for (int j = 1; j <= p.length(); j++) {
+          if (p.charAt(j - 1) == '.' || p.charAt(j - 1) == s.charAt(i - 1)) {
+            dp[i][j] = dp[i - 1][j - 1];
+          } else if (p.charAt(j - 1) == '*') {
+            if (p.charAt(j - 2) != s.charAt(i - 1) && p.charAt(j - 2) != '.') {
+              dp[i][j] = dp[i][j - 2];
+            } else {
+              dp[i][j] = (dp[i][j - 2] || dp[i][j - 1] || dp[i - 1][j]);
+            }
+          }
+        }
+      }
+
+      return dp[s.length()][p.length()];
+    }
+}
+
+```
+### 其他优秀解答
+```
+暂无
+```

daily/README.md

@@ -25,10 +25,16 @@ tag: `Array`
 
 #### [347.Top K Frequent Elements](./2019-06-08.md)
 
-tag: `HashTable Heap`
+tag: `HashTable` `Heap`
 
 时间: 2019-06-08
 
+#### [10.Regular Expression Matching](./2019-06-09.md)
+
+tag:  `String` `Dynamic Programming` `Backtracking`
+
+时间: 2019-06-09
+
 #### [617. Merge Two Binary Trees](./2019-06-10.md)
 
 tag: `Tree`