Wildcard Matching
Total Accepted: 55214 Total Submissions: 317630 Difficulty: Hard
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
- 与regular expression matching相类似
- 找dp的规律就可以
- 主要注意的是初始化,可能会遇到"" 与 "**"对应的情况
- 0(n2)
public class Solution {
public boolean isMatch(String s, String p) {
if ((s == null && p == null) || (s.equals("") && p.equals(""))) {
return true;
}
if ((s == null || p == null || p.equals(""))) {
return false;
}
int m = s.length();
int n = p.length();
boolean[][] dp = new boolean[m + 1][n + 1];
dp[0][0] = true;
int x = 0;
while (x < n && p.charAt(x) == '*') {
for (int i = 0; i <= m; i++) {
dp[i][x + 1] = true;
}
x++;
}
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
char schar = s.charAt(i - 1);
char pchar = p.charAt(j - 1);
if (schar == pchar || pchar == '?') {
dp[i][j] = dp[i - 1][j - 1];
}
if (pchar == '*') {
dp[i][j] = dp[i - 1][j] || dp[i][j - 1];
}
}
}
return dp[m][n];
}
}