[leetcode] 44. Wildcard Matching 解题报告

题目链接: https://leetcode.com/problems/wildcard-matching/

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") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "*") → trueisMatch("aa", "a*") → trueisMatch("ab", "?*") → trueisMatch("aab", "c*a*b") → false

思路: 有两种特殊字符, '?'代表匹配一个任意字符, 这个比较好处理. '*'匹配任意多个字符, 这个需要考虑匹配多个字符. 因此我们可以记下最后出现'*'的位置, 这样当后面位置不匹配的时候再回到这里将不匹配的字符用'*'来匹配. 这样最后再判断是否p指针停留的位置到最后都是*, 如果是的话则可以匹配, 否则不可以匹配. 一个例子如: s = "aa", p = "aa****".

代码如下:

class Solution {public:    bool isMatch(string s, string p) {        int preS=-1, preP=-1, i=0, j=0, len1=s.size(), len2=p.size();        while(i < len1)        {            if(s[i]==p[j] || p[j]=='?') i++, j++;            else if(p[j]=='*') preS=i+1, preP = j++;            else if(preP==-1) return false;            else i = preS, j = preP;        }        while(p[j]=='*') j++;        return i==len1&&j==len2;    }};

参考: https://leetcode.com/discuss/49254/fastest-non-dp-solution-with-o-1-space