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package class12;
// 测试链接 : https://leetcode.com/problems/regular-expression-matching/
public class Code04_RegularExpressionMatch {
public static boolean isValid(char[] s, char[] e) {
// s中不能有'.' or '*'
for (int i = 0; i < s.length; i++) {
if (s[i] == '*' || s[i] == '.') {
return false;
}
}
// 开头的e[0]不能是'*',没有相邻的'*'
for (int i = 0; i < e.length; i++) {
if (e[i] == '*' && (i == 0 || e[i - 1] == '*')) {
return false;
}
}
return true;
}
// 初始尝试版本,不包含斜率优化
public static boolean isMatch1(String str, String exp) {
if (str == null || exp == null) {
return false;
}
char[] s = str.toCharArray();
char[] e = exp.toCharArray();
return isValid(s, e) && process(s, e, 0, 0);
}
// str[si.....] 能不能被 exp[ei.....]配出来! true false
public static boolean process(char[] s, char[] e, int si, int ei) {
if (ei == e.length) { // exp 没了 str
return si == s.length;
}
// exp[ei]还有字符
// ei + 1位置的字符不是*
if (ei + 1 == e.length || e[ei + 1] != '*') {
// ei + 1 不是*
// str[si] 必须和 exp[ei] 能配上!
return si != s.length && (e[ei] == s[si] || e[ei] == '.') && process(s, e, si + 1, ei + 1);
}
// exp[ei]还有字符
// ei + 1位置的字符是*!
while (si != s.length && (e[ei] == s[si] || e[ei] == '.')) {
if (process(s, e, si, ei + 2)) {
return true;
}
si++;
}
return process(s, e, si, ei + 2);
}
// 改记忆化搜索+斜率优化
public static boolean isMatch2(String str, String exp) {
if (str == null || exp == null) {
return false;
}
char[] s = str.toCharArray();
char[] e = exp.toCharArray();
if (!isValid(s, e)) {
return false;
}
int[][] dp = new int[s.length + 1][e.length + 1];
// dp[i][j] = 0, 没算过!
// dp[i][j] = -1 算过返回值是false
// dp[i][j] = 1 算过返回值是true
return isValid(s, e) && process2(s, e, 0, 0, dp);
}
public static boolean process2(char[] s, char[] e, int si, int ei, int[][] dp) {
if (dp[si][ei] != 0) {
return dp[si][ei] == 1;
}
boolean ans = false;
if (ei == e.length) {
ans = si == s.length;
} else {
if (ei + 1 == e.length || e[ei + 1] != '*') {
ans = si != s.length && (e[ei] == s[si] || e[ei] == '.') && process2(s, e, si + 1, ei + 1, dp);
} else {
if (si == s.length) { // ei ei+1 *
ans = process2(s, e, si, ei + 2, dp);
} else { // si没结束
if (s[si] != e[ei] && e[ei] != '.') {
ans = process2(s, e, si, ei + 2, dp);
} else { // s[si] 可以和 e[ei]配上
ans = process2(s, e, si, ei + 2, dp) || process2(s, e, si + 1, ei, dp);
}
}
}
}
dp[si][ei] = ans ? 1 : -1;
return ans;
}
// 动态规划版本 + 斜率优化
public static boolean isMatch3(String str, String pattern) {
if (str == null || pattern == null) {
return false;
}
char[] s = str.toCharArray();
char[] p = pattern.toCharArray();
if (!isValid(s, p)) {
return false;
}
int N = s.length;
int M = p.length;
boolean[][] dp = new boolean[N + 1][M + 1];
dp[N][M] = true;
for (int j = M - 1; j >= 0; j--) {
dp[N][j] = (j + 1 < M && p[j + 1] == '*') && dp[N][j + 2];
}
// dp[0..N-2][M-1]都等于false只有dp[N-1][M-1]需要讨论
if (N > 0 && M > 0) {
dp[N - 1][M - 1] = (s[N - 1] == p[M - 1] || p[M - 1] == '.');
}
for (int i = N - 1; i >= 0; i--) {
for (int j = M - 2; j >= 0; j--) {
if (p[j + 1] != '*') {
dp[i][j] = ((s[i] == p[j]) || (p[j] == '.')) && dp[i + 1][j + 1];
} else {
if ((s[i] == p[j] || p[j] == '.') && dp[i + 1][j]) {
dp[i][j] = true;
} else {
dp[i][j] = dp[i][j + 2];
}
}
}
}
return dp[0][0];
}
}