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") → 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
分析
跟上一题很类似。
主要是'*'的匹配问题。p每遇到一个'*'，就保留住当前'*'的坐标和s的坐标，然后s从前往后扫描，如果不成功，则s++，重新扫描。
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();
}
}