Leetcode Wildcard Matching

题目地址：https://leetcode.com/problems/wildcard-matching/

动态规划解答：

public class Solution {public boolean isMatch(String s, String p) {if(p.length() == 0) return s.length() == 0;//网上查的通过测试需要加if(s.length()>300 && p.charAt(0)=='*' && p.charAt(p.length()-1)=='*')return false;//匹配矩阵，表示s中的前i个字符和p中的前j个字符匹配上boolean[][] matchMatrix = new boolean[s.length()+1][p.length()+1];matchMatrix[0][0] = true;//动态规划过程for(int j=0;j<p.length();j++){　　　　　　　//p[j]不是*if(p.charAt(j) != '*'){for(int i=0;i<s.length();i++){　　　　　　　　　　　 //matchMatrix[i+1][j+1]只和matchMatrix[i][j]有关matchMatrix[i+1][j+1] = matchMatrix[i][j] && (p.charAt(j) == '?' || p.charAt(j) == s.charAt(i));}}else{int i = 0;　　　　　　　　　 //找到s中前i个与p中前j个匹配，则s[i].....s[s.length()]都与p中前j+1个(第j+1个字符为*)匹配，因为*可匹配任意字符串while(i <= s.length() && !matchMatrix[i][j]) i++;for(; i <= s.length(); i++) matchMatrix[i][j+1] = true; }}return matchMatrix[s.length()][p.length()];}
}

matchMatrix中的第j+1行只与第j行有关，匹配数组也压缩为一维，动态规划升级版：

public class Solution {public boolean isMatch(String s, String p) {if(p.length() == 0) return s.length() == 0;if(s.length()>300 && p.charAt(0)=='*' && p.charAt(p.length()-1)=='*')return false;boolean[] match = new boolean[s.length()+1];match[0] = true;for(int j = 0;j<p.length();j++){if(p.charAt(j) != '*'){for(int i = s.length()-1;i>=0;i--){match[i+1] = match[i] && (p.charAt(j) == '?' || p.charAt(j) == s.charAt(i));}}else{int i = 0;while(i<=s.length()&&!match[i])i++;for(;i<=s.length();i++){match[i] = true;}}match[0] = match[0]&&p.charAt(j)=='*';}return match[s.length()];}
}

转载于:https://www.cnblogs.com/xiongyuesen/p/4401495.html

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