[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") → 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

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 si = 0, pi = 0, oldsi = -1, oldpi=-1;
        while(si < s.size())
        {
            if(s[si] == p[pi] || p[pi]=='?')
                si++, pi++;
            else if(pi < p.size() && p[pi] =='*')
            {
                oldsi = si + 1;
                oldpi = pi;
                pi++;
            }
            else if(oldpi!= -1 && s[si] != p[pi])
            {
                si = oldsi;
                pi = oldpi;
            }
            else return false;
        }
        while(pi < p.size() && p[pi]=='*') pi++;
        return pi == p.size();
    }
};

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