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
https://oj.leetcode.com/problems/wildcard-matching/
思路1：二维动归（提交竟然空间超出错误）。 dp[i][j]代表p[0...i]是否匹配 s[0...j]。
- 如果p[i]！='*'，t[i][j] == true 当 t[i-1][j-1]==true &&（p[i]==s[j]||p[i]='.'）
- 如果p[i]=='*'，t[i][j]== true 当 其中一个m使得 t[i-1][m]==true，where 0<=m<j.
思路2：二维状态压缩为一维，原理是一样的。
思路1和思路2代码：
/**
* http://www.darrensunny.me/leetcode-wildcard-matching-2/
* http://www.wangqifox.cn/wordpress/?p=396
*
* @author Dong Jiang
*
*/
public class Solution {
// MLE, the states can be compressed.
public boolean isMatchOld(String s, String p) {
if (s == null)
return p == null;
if (p == null)
return s == null;
int m = p.length();
int n = s.length();
boolean[][] dp = new boolean[m + 1][n + 1];
dp[0][0] = true;
for (int i = 1; i <= m; i++) {
if (p.charAt(i - 1) == '*') {
int j = n + 1;
for (int k = 0; k <= n; k++) {
if (dp[i - 1][k])
j = k;
}
for (; j <= n; j++)
dp[i][j] = true;
} else {
for (int j = 1; j <= n; j++) {
if (dp[i - 1][j - 1] && (p.charAt(i - 1) == s.charAt(j - 1) || p.charAt(i - 1) == '?'))
dp[i][j] = true;
}
}
dp[i][0] = dp[i - 1][0] && (p.charAt(i - 1) == '*');
}
return dp[m][n];
}
public boolean isMatch(String s, String p) {
if (s == null)
return p == null;
if (p == null)
return s == null;
int m = p.length();
int n = s.length();
int count = 0;
for (int i = 0; i < m; i++) {
if (p.charAt(i) != '*')
count++;
}
if (count > n)
return false;
boolean[] dp = new boolean[n + 1];
dp[0] = true;
for (int i = 1; i <= m; i++) {
if (p.charAt(i - 1) == '*') {
int j = n+1;
for (int k = 0; k <= n; k++) {
if (dp[k]){
j = k;
break;
}
}
for (; j <= n; j++)
dp[j] = true;
} else {
for (int j = n; j >= 1; j--) {
dp[j]= (dp[j - 1] && (p.charAt(i - 1) == s.charAt(j - 1) || p.charAt(i - 1) == '?'));
}
}
dp[0] = dp[0] && (p.charAt(i - 1) == '*');
}
return dp[n];
}
public static void main(String[] args) {
System.out.println(new Solution().isMatch("ab", "*a"));
}
}
第二遍记录：注意有一行，因为一维是与上一次循环公用数组（意味着默认不一定是false了），所以不仅要赋值true的情况，还要赋值false的情况。
public class Solution { public boolean isMatch(String s, String p) { if (s == null) return p == null; if (p == null) return s == null; int m = p.length(); int n = s.length(); // to deal with the hardest case... int count = 0; for (int i = 0; i < m; i++) { if (p.charAt(i) != '*') count++; } if (count > n) return false; boolean[] dp = new boolean[n + 1]; dp[0] = true; for (int i = 1; i <= m; i++) { if (p.charAt(i - 1) == '*') { int j = n + 1; for (int k = 0; k <= n; k++) { if (dp[k]) { j = k; break; } } for (; j <= n; j++) dp[j] = true; } else { for (int j = n; j >= 1; j--) { //be careful of this line dp[j]= (dp[j - 1] && (p.charAt(i - 1) == s.charAt(j - 1) || p.charAt(i - 1) == '?')); } } dp[0] = dp[0] && (p.charAt(i - 1) == '*'); } return dp[n]; } }
参考：