LeetCoder_____Regular Expression Matching(10)
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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") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "a*") → trueisMatch("aa", ".*") → trueisMatch("ab", ".*") → trueisMatch("aab", "c*a*b") → true
题意:
模拟正则表达式,给出一个字符串s,和正则表达式p,返回s和p是否匹配。
'.' 可以匹配一个任意字符
‘*’可以匹配零个或者多个前一字符,例如:“q*” 可匹配任意个字符。
例如:
isMatch("aa","a") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "a*") → trueisMatch("aa", ".*") → trueisMatch("ab", ".*") → trueisMatch("aab", "c*a*b") → true
分析:
很显然用递归可以来匹配,但是这个题目要防止超时。可以通过加记忆的话可以减少时间复杂度。
1.如果p为空,那么当s为空就返回true,否则返回false。
2.当p[1]不为*的时候,那么s的第一个字符一定和p的第一个字符匹配或者p的第一个字符为'.'。然后继续匹配剩下的。
3.当p[1]为*的时候,要么p前两个匹配0个也就是s和p.substr(2)匹配,否则那么s的第一个字符一定和p的第一个字符匹配或者p的第一个字符为'.'。然后继续匹配剩下的。
以上是递归匹配的思路,但是如果不加以剪枝可能会超时。
更好的方法是使用动态规划。
转换方程:
if (p[j - 1] != '*')
f[i][j] = f[i - 1][j - 1] && (s[i - 1] == p[j - 1] || '.' == p[j - 1]);
else
f[i][j] = f[i][j - 2] || (s[i - 1] == p[j - 2] || '.' == p[j - 2]) && f[i - 1][j];
代码:
//加记忆的递归class Solution {public: int dp[1002][1002]; Solution(){ memset(dp,-1,sizeof(dp)); } bool isMatch(string s, string p) { if(dp[s.length()][p.length()]!=-1) return (bool)dp[s.length()][p.length()]; 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 dp[s.length()][p.length()]=(isMatch(s, p.substr(2)) || !s.empty() && (s[0] == p[0] || '.' == p[0]) && isMatch(s.substr(1), p)); else return dp[s.length()][p.length()]=(!s.empty() && (s[0] == p[0] || '.' == p[0]) && isMatch(s.substr(1), p.substr(1))); }};
//DP动态规划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<vector<bool>> f(m + 1, vector<bool>(n + 1, false)); 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]; return f[m][n]; }};
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