字符串匹配
作者:互联网
44. Wildcard Matching
Hard
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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
class Solution {
public:
bool isMatch(string s, string p) {
int SIndex = 0,PIndex = 0,SStart = 0,PStart = 0;
while(SIndex < s.length()){
//--s和p一一匹配
if(s[SIndex] == p[PIndex] || p[PIndex] == '?'){
SIndex++;PIndex++;
}
//--遇到通配符*
else if(p[PIndex] == '*'){
PStart = ++PIndex;
SStart = SIndex;
}
//--s[SIndex] != p[PIndex],并且s[SIndex]前面有通配符*
else if(PStart > 0){
PIndex = PStart;
SIndex = ++ SStart;
}
//--s[SIndex] != p[PIndex],并且s[SIndex]前面没有通配符*
else{
return false;
}
}
while(PIndex < p.length() && p[PIndex] == '*'){
PIndex++;
}
if(PIndex == p.length()){
return true;
}
else{
return false;
}
}
};
标签:SIndex,匹配,string,PIndex,Output,字符串,Input,false 来源: https://blog.csdn.net/qq_35747066/article/details/99671406