Given an input string (s) and a pattern (p), 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).
Note:
scould be empty and contains only lowercase lettersa-z.pcould be empty and contains only lowercase lettersa-z, and characters like.or*.
Example 1:
Input: s = "aa" p = "a" Output: false Explanation: "a" does not match the entire string "aa".
Example 2:
Input: s = "aa" p = "a*" Output: true Explanation: '*' means zero or more of the preceding element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".
Example 3:
Input: s = "ab" p = ".*" Output: true Explanation: ".*" means "zero or more (*) of any character (.)".
Example 4:
Input: s = "aab" p = "c*a*b" Output: true Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore, it matches "aab".
Example 5:
Input: s = "mississippi" p = "mis*is*p*." Output: false
思路一：递归思路，先匹配第一个字符。终止条件：p到达末尾而s没有达到，返回false。
dfs(i, j)表示从s的第i个字符和p的第j个字符匹配。
Top-down
1 class Solution {
2 public:
3 bool isMatch(string s, string p) {
4 if (p.empty())
5 return s.empty();
6 bool first_match = (!s.empty() && (s[0] == p[0] || p[0] == '.')); //判断s和p的第一个字符是否匹配
7 if (p.length() >= 2 && p[1] == '*') { //如果p前两个字符是[a-z|.]*,则选择s的第一个字符是否与其匹配
8 return isMatch(s, p.substr(2)) || (first_match && isMatch(s.substr(1), p)); //前面那种情况选择直接跳过
9 } else {
10 return first_match && isMatch(s.substr(1), p.substr(1)); //p首字符不是[a-z|.]*这种情况
11 }
12 }
13 };
思路二：记忆化搜索
1 class Solution {
2 public:
3 bool isMatch(string s, string p) {
4 int lens = s.length(), lenp = p.length();
5 vector<vector<int> > memo(lens + 1, vector<int>(lenp + 1, -1));
6 return isMatch(s, p, 0, 0, memo);
7 }
8 bool isMatch(string s, string p, int i, int j, vector<vector<int> > &memo) {
9 if (memo[i][j] != -1) {
10 return memo[i][j];
11 }
12 int ans;
13 if (j == p.length()) {
14 ans = (i == s.length());
15 } else {
16 bool first_match = (i < s.length() && (p[j] == '.' || s[i] == p[j]));
17 if (j + 1 < p.length() && p[j + 1] == '*') {
18 ans = isMatch(s, p, i, j + 2, memo) || (first_match && isMatch(s, p, i + 1, j, memo));
19 } else {
20 ans = first_match && isMatch(s, p, i + 1, j + 1, memo);
21 }
22 }
23 memo[i][j] = ans;
24 return ans;
25 }
26 };
思路三：动态规划
基于思路二，转化成迭代形式。dp[i][j]表示从下标i开始的s子串与从下标j开始的p子串是否匹配
1 class Solution {
2 public:
3 bool isMatch(string s, string p) {
4 int lens = s.length(), lenp = p.length();
5 vector<vector<bool> > dp(lens + 1, vector<bool>(lenp + 1, false));
6 dp[lens][lenp] = true;
7 for (int i = lens; i >= 0; i--) {
8 for (int j = lenp - 1; j >= 0; j--) {
9 bool first_match = (i < lens) && (p[j] == '.' || (s[i] == p[j]));
10 if (j + 1 < lenp && p[j + 1] == '*') {
11 dp[i][j] = dp[i][j + 2] || (first_match && dp[i + 1][j]);
12 } else {
13 dp[i][j] = first_match && dp[i + 1][j + 1];
14 }
15 }
16 }
17 return dp[0][0];
18 }
19 };
还有另一种形式的dp，dp[i][j]表示长度为i的s和长度为j的p是否匹配。
1 class Solution {
2 public:
3 bool isMatch(string s, string p) {
4 int lens = s.length(), lenp = p.length();
5 vector<vector<bool> > dp(lens + 1, vector<bool>(lenp + 1, false));
6 dp[0][0] = true;
7 for (int i = 0; i <= lens; i++) {
8 for (int j = 1; j <= lenp; j++) {
9 if (p[j - 1] == '*') {
10 dp[i][j] = dp[i][j - 2] || (i > 0 && (s[i - 1] == p[j - 2] || p[j - 2] == '.') && dp[i - 1][j]);
11 } else {
12 dp[i][j] = i > 0 && (s[i - 1] == p[j - 1] || p[j - 1] == '.') && dp[i - 1][j - 1];
13 }
14 }
15 }
16 return dp[lens][lenp];
17 }
18 };