LeetCode 010 Regular Expression Matching

xiaoxiao2021-02-28  5

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

This Solution use 2D DP. beat 90% solutions, very simple.

Here are some conditions to figure out, then the logic can be very straightforward.

1, If p.charAt(j) == s.charAt(i) : dp[i][j] = dp[i-1][j-1]; 2, If p.charAt(j) == '.' : dp[i][j] = dp[i-1][j-1]; 3, If p.charAt(j) == '*': here are two sub conditions: 1 if p.charAt(j-1) != s.charAt(i) : dp[i][j] = dp[i][j-2] //in this case, a* only counts as empty 2 if p.charAt(i-1) == s.charAt(i) or p.charAt(i-1) == '.': dp[i][j] = dp[i-1][j] //in this case, a* counts as multiple a or dp[i][j] = dp[i][j-1] // in this case, a* counts as single a or dp[i][j] = dp[i][j-2] // in this case, a* counts as empty

class Solution { public boolean isMatch(String s, String p) { if(s == null && p == null) { return false; } int lens = s.length(), lenp = p.length(); boolean dp[][] = new boolean[lens + 1][lenp + 1]; // 初始化dp[0][j]，但不需要初始化dp[i][0] dp[0][0] = true; for(int j = 1; j < lenp; j++) { if(p.charAt(j) == '*' && dp[0][j - 1]) { dp[0][j + 1] = true; } } for(int i = 0; i < lens; i++) { for(int j = 0; j < lenp; j++) { if(p.charAt(j) == '.' || s.charAt(i) == p.charAt(j)) { dp[i + 1][j + 1] = dp[i][j]; } if(p.charAt(j) == '*') { if(p.charAt(j - 1) != s.charAt(i) && p.charAt(j - 1) != '.') { dp[i + 1][j + 1] = dp[i + 1][j - 1]; } else { dp[i + 1][j + 1] = (dp[i][j + 1] || dp[i + 1][j] || dp[i + 1][j - 1]); } } } } return dp[lens][lenp]; } }

dp[i + 1][j + 1]代表s串前i个与p串前j个能否匹配。