Leetcode 10. Regular Expression Matching (Hard) (cpp)

最新推荐文章于 2019-10-27 16:29:57 发布

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本文介绍了一种解决LeetCode 10题目的方法：实现支持'.'和'*'的正则表达式匹配。通过动态规划算法，文章详细解释了如何判断一个字符串是否完全匹配给定的正则表达式。

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Leetcode 10. Regular Expression Matching (Hard) (cpp)

Tag: Dynamic Programming, Backtracking, String

Difficulty: Hard

/*

10. Regular Expression Matching (Hard)

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

*/
class Solution {
public:
	bool isMatch(string s, string p) {
		int m = s.size(), n = p.size();
		vector<vector<bool>> t(m + 1, vector<bool>(n + 1, false));
		t[0][0] = true;
		for (int i = 1; i <= m; i++) {
			t[i][0] = false;
		}
		for (int j = 1; j <= n; j++) {
			t[0][j] = j > 1 && '*' == p[j - 1] && t[0][j - 2];
		}
		for (int i = 1; i <= m; i++) {
			for (int j = 1; j <= n; j++) {
				if (p[j - 1] != '*') {
					t[i][j] = t[i - 1][j - 1] && (s[i - 1] == p[j - 1] || '.' == p[j - 1]);
				}
				else {
					t[i][j] = t[i][j - 2] || (s[i - 1] == p[j - 2] || '.' == p[j - 2]) && t[i - 1][j];
				}
			}
		}
		return t[m][n];
	}
};