leetcode 44. Wildcard Matching
作者:互联网
Given an input string (s
) and a pattern (p
), 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).
Note:
s
could be empty and contains only lowercase lettersa-z
.p
could be empty and contains only lowercase lettersa-z
, and characters like?
or*
.
Example 1:
Input: s = "aa" p = "a" Output: false Explanation: "a" does not match the entire string "aa".
Example 2:
Input: s = "aa" p = "*" Output: true Explanation: '*' matches any sequence.
Example 3:
Input: s = "cb" p = "?a" Output: false Explanation: '?' matches 'c', but the second letter is 'a', which does not match 'b'.
Example 4:
Input: s = "adceb" p = "*a*b" Output: true Explanation: The first '*' matches the empty sequence, while the second '*' matches the substring "dce".
Example 5:
Input: s = "acdcb" p = "a*c?b" Output: false
思路:基本和leetcode10 正则表达式匹配思路一致。
方法一:记忆化搜索
1 class Solution { 2 public: 3 bool isMatch(string s, string p) { 4 int lens = s.length(), lenp = p.length(); 5 vector<vector<int> > memo(lens + 1, vector<int>(lenp + 1, -1)); 6 return isMatch(s, p, 0, 0, memo); 7 } 8 bool isMatch(string s, string p, int i, int j, vector<vector<int> > &memo) { 9 if (memo[i][j] != -1) { 10 return memo[i][j]; 11 } 12 int ans; 13 if (j >= p.length()) { 14 ans = (i == s.length()); 15 } else { 16 bool first_match = (i < s.length() && (s[i] == p[j] || p[j] == '?' || p[j] == '*')); 17 if (p[j] == '*') { 18 ans = isMatch(s, p, i, j + 1, memo) || (first_match && isMatch(s, p, i + 1, j, memo)); 19 } else { 20 ans = first_match && isMatch(s, p, i + 1, j + 1, memo); 21 } 22 } 23 memo[i][j] = ans; 24 return ans; 25 } 26 };
方法二:动态规划
1 class Solution { 2 public: 3 bool isMatch(string s, string p) { 4 int lens = s.length(), lenp = p.length(); 5 bool dp[lens + 1][lenp + 1]; 6 memset(dp, false, sizeof(dp)); 7 //vector<vector<bool> > dp(lens + 1, vector<bool>(lenp + 1, false)); //用vector耗时 8 dp[0][0] = true; 9 for(int i = 1; i <= lenp; i ++) { 10 dp[0][i] = p[i - 1] == '*' && dp[0][i - 1]; 11 } 12 for(int i = 1; i <= lens; i ++) { 13 for(int j = 1; j <= lenp; j ++) { 14 if(p[j - 1] != '*') { 15 if ((s[i - 1] == p[j - 1]) || (p[j - 1] == '?')) { 16 dp[i][j] = dp[i - 1][j - 1]; 17 } 18 } else { 19 dp[i][j] = (dp[i][j - 1] || dp[i - 1][j]); //如果长度为j的p,最后一个是'*',可以将'*'变成s的最后一个字符,则有dp[i][j] = dp[i - 1][j], 或者'*'变为空。dp[i][j] = dp[i][j - 1] 20 } 21 } 22 } 23 return dp[lens][lenp]; 24 } 25 };
标签:string,int,44,leetcode,length,vector,isMatch,Wildcard,memo 来源: https://www.cnblogs.com/qinduanyinghua/p/11528190.html