description:
匹配字符串，具体看例子
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 letters a-z.
p could be empty and contains only lowercase letters a-z, and characters like ? or *.
Example:
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
answer:
class Solution {
public:
bool isMatch(string s, string p) {
int i = 0, j = 0, iStar = -1, jStar = -1;
while (i < s.size()) {
if (s[i] == p[j] || p[j] == '?') {
++i; ++j;
} else if (p[j] == '*') {
iStar = i;
jStar = j++; // 这里是先标记位置，然后在挪动 j index
} else if (iStar >= 0) {
i = ++iStar; //这里一定是++iStar,而不是 iStar+1，因为不但要挪，还要改变 iStar
j = jStar + 1;
} else return false;
}
while (j < p.size() && p[j] == '*') ++j;
return j == p.size();
}
};
relative point get√：
hint :
两个都从头开始走，一样就继续，不一样就三种可能，一种是‘?’，继续；一种是'*',标记完位置之后上边字符串不变，下边走一步（先假设'*'是空，然后往后走，走到不行了再用这个开挂神器）；最后一种就是真的不一样，这个时候就想起来看之前有没有'*'，有的话，就用'*'去匹配上边的，一直匹配到上边出现某个字符和下边‘*’右边那个字符一样，然后继续开始之前正常的运作。最后上边的字符都匹配完了，如果下面字符还剩下了，还不是‘*’，那也是失败。