LeetCode: 44. Wildcard Matching

题目描述

Implement wildcard pattern matching with support for ‘?’ and ‘*’.

‘?’ Matches any single character.
‘*’ Matches any sequence of characters (including the empty sequence).

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”, “*”) → true
isMatch(“aa”, “a*”) → true
isMatch(“ab”, “?*”) → true
isMatch(“aab”, “c*a*b”) → false

题目大意：给定模式串 p, 匹配串 s。其中，p含有通配符 (Wildcard)。 ‘?’ 代表可以匹配任意一个字符， ‘*’ 代表可以匹配任意个数任意字符。判断 p 串和 s 串是否匹配。

解题思路 — 动态规划

记s串的前i个字符和p串的前j个字符能否匹配为 dp[i][j]。
如果模式串p和匹配串s都为空，则匹配成功， dp[0][0] = true;。
如果模式串p为空，s串不为空，则匹配失败， dp[i][0] = false, 其中 0 < i <= s.size()。
如果模式串p不为空， s串为空，i > 0, 若 p[i-1] == '*' 则 dp[0][i] 的真值受 dp[0][i-1] 的影响。否则， dp[0][i] = false 。即，
当p串和s串都不为空时，状态转移方程如下，

AC代码

class Solution {
public:bool isMatch(string s, string p) {vector<vector<bool>> dp;dp.resize(s.size()+1);for(auto& vb : dp) vb.resize(p.size()+1, false);dp[0][0] = true;for(int i = 1; i <= p.size(); ++i){if(p[i-1] == '*') dp[0][i] = dp[0][i-1];}for(int i = 1; i <= s.size(); ++i){for(int j = 1; j <= p.size(); ++j){if(s[i-1] == p[j-1] || p[j-1] == '?'){dp[i][j] = dp[i-1][j-1];}else if(p[j-1] == '*'){for(int k = 0; k <= i; ++k){if(dp[k][j-1] == true){dp[i][j] = true;break;}}}else{dp[i][j] = false;}}}return dp[s.size()][p.size()];}
};

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LeetCode: 44. Wildcard Matching

LeetCode: 44. Wildcard Matching

题目描述

解题思路 — 动态规划

AC代码

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