LeetCode-44-Wildcard Matching
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算法描述:
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
解题思路:动态规划题。递推式:
dp[i][j] = dp[i-1][j-1] if(s[i-1]=p[j-1] || p[j-1]==‘?‘)
dp[i][j] = dp[i-1][j] || dp[j][i-1] if(p[j-1]==‘*‘)
注意初始化时,p[j-1] =‘*‘的处理。
bool isMatch(string s, string p) { int m = s.size(); int n = p.size(); vector<vector<bool>> dp(m+1,vector<bool>(n+1,false)); dp[0][0] = true; for(int i=1; i<=n; i++){ if(p[i-1]==‘*‘){ dp[0][i] = dp[0][i-1]; } } for(int i=1; i <=m; i++){ for(int j=1; j<=n; j++){ if(p[j-1]==‘*‘){ dp[i][j] = dp[i-1][j] || dp[i][j-1] ; } if(s[i-1]==p[j-1] || p[j-1]==‘?‘){ dp[i][j] = dp[i-1][j-1]; } } } return dp[m][n]; }
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