LeetCode题解——10Regular Expression Matching

题目：

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

这种Assignment的问题，显然可以用动态规划解决。

假设f[i][j]=true 表示s[0]....s[i-1]和p[0]...p[j-1] Match成功；那么判断f[i][j]是否Match可以分情况讨论；1.p[j-1]!='*'，若s[i-1]==p[j-1] && f[i-1][j-1],那么f[i][j]=true;   2.p[j-1]=='*',denote p[j-2]==x;  (1)x* denote empty, f[i][j-2]==ture thenf[i][j] = true;  (2)x* denote x*x, s[i-1]==p[j-2] && f[i-1][j]

class Solution {
public:
	bool isMatch(string s, string p) {
		/**
		* f[i][j]: if s[0..i-1] matches p[0..j-1]
		* if p[j - 1] != '*'
		*      f[i][j] = f[i - 1][j - 1] && s[i - 1] == p[j - 1]
		* if p[j - 1] == '*', denote p[j - 2] with x
		*      f[i][j] is true iff any of the following is true
		*      1) "x*" repeats 0 time and matches empty: f[i][j - 2]
		*      2) "x*" repeats >= 1 times and matches "x*x": s[i - 1] == x && f[i - 1][j]
		* '.' matches any single character
		*/
		int m = s.size(), n = p.size();
		vector> f(m + 1, vector(n + 1, false));//vector的初始化

		f[0][0] = true;
		for (int i = 1; i <= m; i++)
			f[i][0] = false;
		// p[0.., j - 3, j - 2, j - 1] matches empty iff p[j - 1] is '*' and p[0..j - 3] matches empty
		for (int j = 1; j <= n; j++)
			f[0][j] = (j > 1 && '*' == p[j - 1] && f[0][j - 2]);

		for (int i = 1; i <= m; i++)
			for (int j = 1; j <= n; j++)
				if (p[j - 1] != '*')
					f[i][j] = f[i - 1][j - 1] && (s[i - 1] == p[j - 1] || '.' == p[j - 1]);
				else
					// p[0] cannot be '*' so no need to check "j > 1" here
					f[i][j] = f[i][j - 2] || (s[i - 1] == p[j - 2] || '.' == p[j - 2]) && f[i - 1][j];

		for(int i = 0; i<=m;i++)
		{
			for(int j=0;j<=n;j++)
				cout<

16ms

用递归

class Solution {
public:
	bool isMatch(string s, string p) {
		if (p.empty())    return s.empty();

		if ('*' == p[1])
			// x* matches empty string or at least one character: x* -> xx*
			// *s is to ensure s is non-empty
			return (isMatch(s, p.substr(2)) || !s.empty() && (s[0] == p[0] || '.' == p[0]) && isMatch(s.substr(1), p));
		else
			return !s.empty() && (s[0] == p[0] || '.' == p[0]) && isMatch(s.substr(1), p.substr(1));
	

648ms