Create 192 Wildcard Matching.java

Author: ChenHCY

Leetcode Question/Hard/192 Wildcard Matching.java

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+/* LintCode 192 · Wildcard Matching
+Description
+Given an input string s and a pattern p, implement wildcard pattern matching with support for '?' and '*'. The matching rules are as follows：
+
+'?' Matches any single character.
+'*' Matches any sequence of characters (including the empty sequence).
+The matching should cover the entire input string (not partial).
+
+Example 1
+Input:
+"aa"
+"a"
+Output: false
+
+Example 2
+Input:
+"aa"
+"aa"
+Output: true
+
+Example 3
+Input:
+"aaa"
+"aa"
+Output: false
+
+Example 4
+Input:
+"aa"
+"*"
+Output: true
+Explanation: '*' can replace any string
+
+Example 5
+Input:
+"aa"
+"a*"
+Output: true
+
+Example 6
+Input:
+"ab"
+"?*"
+Output: true
+Explanation: '?' -> 'a' '*' -> 'b'
+
+Example 7
+Input:
+"aab"
+"c*a*b"
+Output: false
+*/
+
+Problem Solving Step Instructions：
+1, If p.charAt(j-1) == s.charAt(i-1) :  dp[i][j] = dp[i-1][j-1];
+2, If p.charAt(j-1) == '.' : dp[i][j] = dp[i-1][j-1];
+3, If p.charAt(j-1) == '*': 
+   here are two sub conditions:
+               1. * counts as singe time or empty case: dp[i][j] = dp[i][j-1]
+               2. * counts as multiple times: dp[i][j] = dp[i-1][j] 
+
+public class Solution {
+    /**
+     * @param s: A string
+     * @param p: A string includes "?" and "*"
+     * @return: is Match?
+     */
+    public boolean isMatch(String s, String p) {
+        // write your code here
+        /*Dp State: dp[i][j]的含义是s[0-i] 与 p[0-i]是否匹配
+          Dp Initialize: dp[0][0] = true
+          Dp function: 1. s.charAt(i - 1) = p.charAt(j - 1) => dp[i][j] = dp[i-1][j-1]
+                       2. p.charAt(j - 1) = '?'  // '?' Matches any single character.
+                          ==> dp[i][j] = dp[i-1][j-1]
+                       3. p.charAt(j - 1) = '*' // '*' Matches any sequence of characters 
+                          ==> * Matches a single time: dp[i][j] = dp[i][j-1]
+                          ==> * Matches a multiple times: dp[i][j] = dp[i-1][j]
+          Dp Answer: return dp[s.length()][p.length()]
+        */
+
+        if(s == null || p == null){
+            return false;
+        }
+        //Dp State:
+        boolean[][] dp = new boolean[s.length() + 1][p.length() + 1];
+        //Dp Initialize:
+        dp[0][0] = true;
+
+        //Preprocessing functon for this kinds solution: s = "aab" , p ="c*aab" 
+        for(int k = 1; k <= p.length(); k++){
+            if(p.charAt(k - 1) == '*'){
+                dp[0][k] = dp[0][k - 1];
+            }
+        }
+
+        //Dp function:
+        for(int i = 1; i <= s.length(); i++){
+            for(int j = 1; j <= p.length(); j++){
+                //1. s.charAt(i) = p.charAt(j)
+                if(s.charAt(i - 1) == p.charAt(j - 1)){
+                    dp[i][j] = dp[i - 1][j - 1];
+                }
+                //2. p.charAt(j) = '?'
+                if(p.charAt(j - 1) == '?'){
+                    dp[i][j] = dp[i - 1][j - 1];
+                }
+                //3. p.charAt(j) = '*' 
+                if(p.charAt(j - 1) == '*' ){
+                    dp[i][j] = dp[i][j-1] || dp[i-1][j];
+                }
+            }
+        }
+
+        //Dp Answer:
+        return dp[s.length()][p.length()];
+    }
+}