Leetcode 10: Regular Expression Matching
2017-08-16 23:25
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Leetcode 10: Regular Expression Matching
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
分析
其实我老是会忘正则表达式里的一些概念比如 * 是指与前一个字符的匹配情况: 0 1 2 3 … 个
还是 dp,这里确实是涉及子问题的。
dp[i][j] : s[0…i-1] 与 p[0…j-1]的匹配情况,true or false
如何求解dp呢?
dp[i][j] : s[0…i-1] p[0…j-1] , 讨论 p[j-1]
若 p[j-1] 为 *
则是关于 p[j-2] 匹配 0/1/多个的情况了
匹配0个 : s[0…i-1] p[0…j-3]
dp[i][j] = dp[i][j-2]
匹配1个: s[0…i-1] p[0…j-2]
dp[i][j] = dp[i][j-1]
匹配多个: s[0…i-2] p[0…j-1]
dp[i][j] = dp[i-1][j-1] && (p[j-2] == s[i-1] || p[j-2] == '.')
若p[j-1] 不为 *
dp[i][j] = dp[i-1][j-1] && (p[j-1]==s[i-1] || p[j-1]=='.')
dp的初始化也值得探讨一下:
int[][] dp = new int[length][length]; dp[0][0] = true; //dp[0][j], 考虑p形如 x* 格式的字符串,需要想一想 for(int j=1; j<=p.length(); j++){ if(p.charAt(j-1)=='*' && dp[0][j-2]){ dp[0][j] = true; } }
解答
public class Solution { public boolean isMatch(String s, String p) { return isMatchByDP(s,p); } /** * dp[i][j] saves the s[0:i-1] and p[0:j-1] match result. */ public boolean isMatchByDP(String s, String p){ boolean[][] dp = new boolean[s.length()+1][p.length()+1]; dp[0][0]=true; // init dp[0][j], null character match x*, only match zero. for(int j=1; 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)!='*'){ dp[i][j]= dp[i-1][j-1] &&(s.charAt(i-1)==p.charAt(j-1) || p.charAt(j-1)=='.'); } else{ // p[j-1]=='*' //match one p[j-2]match zero p[j-2] if(dp[i][j-1] || dp[i][j-2] || (dp[i-1][j] && (s.charAt(i-1)==p.charAt(j-2)||p.charAt(j-2)=='.'))){ dp[i][j]=true; } else { dp[i][j]=false; } } } } return dp[s.length()][p.length()]; } }
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