标签：style blog http color io strong for art
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). 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", "a*") → true isMatch("aa", ".*") → true isMatch("ab", ".*") → true isMatch("aab", "c*a*b") → true
算法思路：
二维动态规划，dp[i + 1][j + 1]表示字符串s(0~ i )和p(0~j)的匹配情况。
初始状态：dp[0][0] = true;
当s[i] == p[j] || p[j] == ‘.‘ 则dp[i][j] = dp[i - 1][j - 1]
当p[j] == ‘*‘时：分两种情况：
1. s[i] != p[j - 2] && p[j - 2] != ‘.‘ 则dp[i][j] = dp[i][j - 2];
2. else dp[i][j] = dp[i][j - 2] | dp[i - 1][j];
代码如下：
1 public class Solution { 2 public boolean isMatch(String s, String p) { 3 int height = s.length(),width = p.length(); 4 boolean[][] dp = new boolean[height + 1][width + 1]; 5 dp[0][0] = true; 6 for(int i = 1; i <= width; i++){ 7 if(p.charAt(i - 1) == ‘*‘) dp[0][i] = dp[0][i - 2]; 8 } 9 for(int i = 1; i <= height; i++){ 10 for(int j = 1; j <= width; j++){ 11 char sChar = s.charAt(i - 1); 12 char pChar = p.charAt(j - 1); 13 if(sChar == pChar || pChar == ‘.‘){ 14 dp[i][j] = dp[i - 1][j - 1]; 15 }else if(pChar == ‘*‘){ 16 if(sChar != p.charAt(j - 2) && p.charAt(j - 2) != ‘.‘){ 17 dp[i][j] = dp[i][j - 2]; 18 }else{ 19 dp[i][j] = dp[i][j - 2] | dp[i - 1][j]; 20 } 21 } 22 } 23 } 24 return dp[height][width]; 25 } 26 }
[leetcode]Regular Expression Matching,布布扣,bubuko.com
[leetcode]Regular Expression Matching
标签：style blog http color io strong for art
原文地址：http://www.cnblogs.com/huntfor/p/3899689.html
