LeetCode 44. Wildcard Matching
2016-10-11 20:25
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题目描述
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).
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”, “*”) → true
isMatch(“aa”, “a*”) → true
isMatch(“ab”, “?*”) → true
isMatch(“aab”, “c*a*b”) → false
题目解析
这里我使用了leetcode10. Regular Expression Matching(hard)的解题思路,逻辑如下:c[i][j] 代表s[:,i]与p[:,j]是否匹配。
p[j]==’*’ then
1.c[i][j]=c[i][j-1]; ‘*’被匹配0次;
2. c[i][j]=c[i-1][j];’*’被匹配1次及以上;
p[j]==’?’then
c[i][j]==c[i-1][j-1];
else
c[i][j]==c[i-1][j-1]&&s[i]==p[j];
但是这里出现了一点问题,运行时间1692ms,排在后5%,一定是有另外的更快的解法。
第一次尝试的代码
class Solution { public: bool isMatch(string s, string p) { int m=s.size(),n=p.size(); vector<vector<bool> >c(m+1,(vector<bool>(n+1,false))); c[0][0]=true; for(int i=0;i<=m;i++) { for(int j=1;j<=n;j++) { if(p[j-1]=='*') { c[i][j]=c[i][j-1]||i>0&&c[i-1][j]; } else if(p[j-1]=='?') { c[i][j]=i>0&&c[i-1][j-1]; } else { c[i][j]=i>0&&s[i-1]==p[j-1]&&c[i-1][j-1]; } } } return c[m] ; } };
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