Leetcode-44. Wildcard Matching

题目介绍

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

递归 （超时）

bool isMatch(string s, string p) {

if(p.empty()){
return s.empty();
}

if(p[0]=='*'){
int i=1;
while(p[i]=='*'){
i++;
}
p = p.substr(i);

if(p==""){
return true;
}

for(int j=0;j<s.length();j++){
if(isMatch(s.substr(j), p)){
return true;
}
}

return false;

}else{
return (p[0]=='?'|| p[0]==s[0])&& !s.empty()&&isMatch(s.substr(1),p.substr(1));
}

}

DP

bool isMatch(string s, string p) {
int m = s.length(),n=p.length();
vector<vector<bool>> d(m+1,vector<bool>(n+1,false));
d[0][0]=true;

//Important
//https://leetcode.com/discuss/54278/my-java-dp-solution-using-2d-table
for(int i=1; i<=n; i++){
if(p[i-1] == '*')
d[0][i] = d[0][i-1];
}

//https://leetcode.com/discuss/43966/accepted-c-dp-solution-with-a-trick
for(int j=1;j<=n;j++){
for(int i=1;i<=m;i++){
if(p[j-1] != '*'){
d[i][j] = (s[i-1] == p[j-1] || p[j-1] == '?') && d[i-1][j-1];
}
else{
d[i][j] = d[i-1][j] || d[i][j-1];
}
}
}

return d[m]
;
}