# Wildcard Matching

### 描述​

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

### 递归版​

// Wildcard Matching// 递归版，会超时，用于帮助理解题意// 时间复杂度O(n!*m!)，空间复杂度O(n)class Solution {    public boolean isMatch(String s, String p) {        return isMatch(s, 0, p, 0);    }    private boolean isMatch(String s, int i, String p, int j) {        if (i == s.length() && j == p.length()) return true;        if (i == s.length() || j == p.length()) return false;        if (p.charAt(j) == '*') {            while (j < p.length() && p.charAt(j) == '*') ++j;  //skip continuous '*'            if (j == p.length()) return true;            while (i < s.length() && !isMatch(s, i, p, j)) ++i;            return i < s.length();        }        else if (p.charAt(j) == s.charAt(i) || p.charAt(j) == '?')            return isMatch(s, ++i, p, ++j);        else return false;    }}

### 迭代版​

// Wildcard Matching// 迭代版，时间复杂度O(n*m)，空间复杂度O(1)public class Solution {    public boolean isMatch(String s, String p) {        int i = 0, j = 0;        int ii = -1, jj = -1;        while (i < s.length()) {            if (j < p.length() && p.charAt(j) == '*') {                // skip continuous '*'                while (j < p.length() && p.charAt(j) == '*') ++j;                if (j == p.length()) return true;                ii = i;                jj = j;            }            if (j < p.length() && (p.charAt(j) == '?' || p.charAt(j) == s.charAt(i))) {                ++i; ++j;            } else {                if (ii == -1) return false;                ++ii;                i = ii;                j = jj;            }        }        // skip continuous '*'        while (j < p.length() && p.charAt(j) == '*') ++j;        return i == s.length() && j == p.length();    }}