标签：style blog http color 使用 io 2014 for
Implement wildcard pattern matching with support for ‘?‘ and ‘*‘.
‘?‘ Matches any single character. ‘*‘ Matches any sequence of characters (including the empty sequence). The matching should cover the entire input string (not partial). The function prototype should be: bool isMatch(const char *s, const char *p) Some examples: isMatch("aa","a") → false isMatch("aa","aa") → true isMatch("aaa","aa") → false isMatch("aa", "*") → true isMatch("aa", "a*") → true isMatch("ab", "?*") → true isMatch("aab", "c*a*b") → false
题解：昨天晚上写了一晚上的递归，一直TLE。早上果断改用dp了，结果dp还TLE了好多次，真是说多了都是泪。
dp的思想很简单：用二维数组dp[s_length+1][p_length+1]记录结果。
开始以为这样就可以过了，事实证明I am too young too simple。这样还是会超时。要优化两个地方：
最后AC的代码如下：
1 public class Solution { 2 public boolean isMatch(String s, String p) { 3 int m = s.length(); 4 int n = p.length(); 5 6 int count = 0; 7 for(int indexP = 0;indexP<n;indexP++) 8 if(p.charAt(indexP) != ‘*‘) 9 count++; 10 if(count > m) 11 return false; 12 13 boolean[][] dp =new boolean[m+1][n+1]; 14 dp[0][0] = true; 15 16 for(int j = 1;j<=n;j++) 17 { 18 char ch_p = p.charAt(j-1); 19 if(ch_p ==‘*‘ && dp[0][j-1]) 20 dp[0][j]= true; 21 22 for(int i= 1;i<=m;i++){ 23 char ch_s = s.charAt(i-1); 24 if(ch_p ==‘*‘) 25 dp[i][j]= dp[i-1][j]|| dp[i][j-1]; 26 else if(ch_p == ‘?‘ || ch_p == ch_s){ 27 dp[i][j] = dp[i-1][j-1]; 28 } 29 else 30 dp[i][j]= false; 31 } 32 } 33 return dp[m][n]; 34 } 35 }
【leetcode刷题笔记】Wildcard Matching,布布扣,bubuko.com
【leetcode刷题笔记】Wildcard Matching
标签：style blog http color 使用 io 2014 for
原文地址：http://www.cnblogs.com/sunshineatnoon/p/3869721.html
