题目
思路
正则匹配
动态规划
同leetcode:10. 正则表达式匹配
题解
package hwod;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class StrMatch {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String[] strArr = sc.nextLine().split(" ");
String p = sc.nextLine();
List<Integer> res = strMatch2(strArr, p);
if (res.size() == 0) {
System.out.println(-1);
return;
}
for (int i = 0; i < res.size(); i++) {
if (i != 0) System.out.print(",");
System.out.print(res.get(i));
}
}
//方案一:正则匹配
private static List<Integer> strMatch(String[] strArr, String p) {
List<Integer> res = new ArrayList<>();
p = "^" + p + "$";
for (int i = 0; i < strArr.length; i++) {
if (strArr[i].matches(p)) {
res.add(i);
}
}
return res;
}
//方案二:动态规划
private static List<Integer> strMatch2(String[] strArr, String p) {
List<Integer> res = new ArrayList<>();
for (int i = 0; i < strArr.length; i++) {
if (checked(strArr[i], p)) {
res.add(i);
}
}
return res;
}
private static boolean checked(String s, String p) {
int m = s.length() + 1, n = p.length() + 1;
int[][] dp = new int[m][n]; //s[:i]和p[:j]匹配
dp[0][0] = 1;
for (int i = 2; i < n; i += 2) {
if (p.charAt(i - 1) == '*') dp[0][i] = dp[0][i - 2];
}
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
if (p.charAt(j - 1) == '*') {
if (dp[i][j - 2] == 1 //a ab*
|| (dp[i - 1][j] == 1 && s.charAt(i - 1) == p.charAt(j - 2)) // abb ab*
|| (dp[i - 1][j] == 1 && p.charAt(j - 2) == '.') // abb a.*
) dp[i][j] = 1;
} else {
if ((dp[i - 1][j - 1] == 1 && p.charAt(j - 1) == '.') // ab a.
|| (dp[i - 1][j - 1] == 1 && p.charAt(j - 1) == s.charAt(i - 1)) //ab ab
) dp[i][j] = 1;
}
}
}
return dp[m - 1][n - 1] == 1;
}
}
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