LeetCode Regular Expression Matching

class Solution {
#define SINGLE 1
#define MULTIP 2
public:
bool isMatch(const char *s, const char *p) {
if (s == NULL || p == NULL) return true;

vector<pair<char, int> > pattern;

int     pos = 0;
char     ch = '\0';

while ((ch = p[pos++]) != '\0') {
if (ch == '*') {
pattern.back().second = MULTIP;
continue;
}
pattern.push_back(make_pair(ch, SINGLE));
}

return isMatch(s, 0, pattern, 0);
}

bool isMatch(const char *s, int pos, vector<pair<char, int> > &pattern, int ppos) {
if (ppos == pattern.size()) {
return s[pos] == '\0';
}
int i = pos;
pair<char, int> &p = pattern[ppos];

int     offset = (p.second == SINGLE) ? 0 : -1;
char     ch = '\0';
while (offset < 0 || (ch = s[pos + offset]) != '\0') {
if (offset >= 0 && !cmp(ch, p.first)) {
return false;
}
if (isMatch(s, pos + offset + 1, pattern, ppos + 1)) {
return true;
}
if (p.second == SINGLE) break;
offset++;
}
return false;
}

bool cmp(const char a, const char b) {
if (a == '.' || b == '.') return true;
return a == b;
}
};

暴力法, 216ms,

改写为记忆搜索,这里假设字符串和模式字符串的长度都不超过unsigned short 范围(6w+)， 84ms

class Solution {
#define SINGLE 1
#define MULTIP 2
private:
unordered_map<unsigned int, bool> memo;
public:
bool isMatch(const char *s, const char *p) {
if (s == NULL || p == NULL) return true;
memo.clear();

vector<pair<char, int> > pattern;

int     pos = 0;
char     ch = '\0';

while ((ch = p[pos++]) != '\0') {
if (ch == '*') {
pattern.back().second = MULTIP;
continue;
}
pattern.push_back(make_pair(ch, SINGLE));
}

return isMatch(s, 0, pattern, 0);
}

bool isMatch(const char *s, int pos, vector<pair<char, int> > &pattern, int ppos) {
unsigned int id = (pos << 16) | ppos;

unordered_map<unsigned int, bool>::iterator iter = memo.find(id);
if (memo.find(id) != memo.end()) return iter->second;

bool res = false;

if (ppos == pattern.size()) {
res = s[pos] == '\0';
memo.insert(make_pair(id, res));
return res;
}
int i = pos;
pair<char, int> &p = pattern[ppos];

int     offset = (p.second == SINGLE) ? 0 : -1;
char     ch = '\0';
while (offset < 0 || (ch = s[pos + offset]) != '\0') {
if (offset >= 0 && !cmp(ch, p.first)) {
res = false;
break;
}
if (isMatch(s, pos + offset + 1, pattern, ppos + 1)) {
res = true;
break;
}
if (p.second == SINGLE) break;
offset++;
}
memo.insert(make_pair(id, res));
return res;
}

bool cmp(const char a, const char b) {
if (a == '.' || b == '.') return true;
return a == b;
}
};

第二轮：

Implement regular expression matching with support for

'.'

and

'*'

.

'.' Matches any single character.
'*' Matches zero or more of the preceding element.

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", "a*") → true
isMatch("aa", ".*") → true
isMatch("ab", ".*") → true
isMatch("aab", "c*a*b") → true

看到这样的题真有点写不下去，还是找了一发资料：

class Solution {
public:
bool isMatch(const char *s, const char *p) {
if (s == NULL || p == NULL) {
return false;
}
if (*p == '\0') {
return *s == '\0';
}

if (*(p+1) != '*') {
return (*p == *s || (*p == '.' && *s != '\0')) && isMatch(s + 1, p + 1);
}

char curpat = *p;
const char* sp = s;

while (*sp == curpat || (curpat == '.' && *sp != '\0')) {
if (isMatch(sp, p + 2)) {
return true;
}
sp++;
}
return isMatch(sp, p + 2);
}
};

在discuss里找到一个dp解法：

class Solution {
public:
bool isMatch(const char *s, const char *p) {
const int slen = strlen(s);
const int plen = strlen(p);
vector<vector<bool> > dp(slen+1, vector<bool>(plen+1, false));

// when src string is empty, pattern string should be in form of
// a*b*c*...
for (int i=2; i<=plen; i+=2) {
if (p[i-1] == '*') {
dp[0][i] = true;
} else {
break;
}
}
dp[0][0] = true;

for (int i=1; i<=slen; i++) {
for (int j=1; j<=plen; j++) {
if (s[i-1] == p[j-1] || p[j-1] == '.') {
dp[i][j] = dp[i-1][j-1];
continue;
}
if (p[j-1] == '*') {
if (p[j-2] == s[i-1] || p[j-2] == '.') {
dp[i][j] = dp[i-1][j] || dp[i][j-2];
} else {
dp[i][j] = dp[i][j-2];
}
}
}
}
return dp[slen][plen];
}
};