题目:
Given an input string ( '.' Matches any single character. '*' Matches zero or more of the preceding element. The matching should cover the entire input string (not partial). Note:
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 precedeng 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 |
给定一个字符串 ( '.' 匹配任意单个字符。 '*' 匹配零个或多个前面的元素。 匹配应该覆盖整个字符串 ( 说明:
示例 1: 输入: s = "aa" p = "a" 输出: false 解释: "a" 无法匹配 "aa" 整个字符串。 示例 2: 输入: s = "aa" p = "a*" 输出: true 解释: '*' 代表可匹配零个或多个前面的元素, 即可以匹配 'a' 。因此, 重复 'a' 一次, 字符串可变为 "aa"。 示例 3: 输入: s = "ab" p = ".*" 输出: true 解释: ".*" 表示可匹配零个或多个('*')任意字符('.')。 示例 4: 输入: s = "aab" p = "c*a*b" 输出: true 解释: 'c' 可以不被重复, 'a' 可以被重复一次。因此可以匹配字符串 "aab"。 示例 5: 输入: s = "mississippi" p = "mis*is*p*." 输出: false |
思路:用递归思路,但是如果写 这种递归,时间不能通过。分为3种情况,1.当模板p为空时,若s也空返回true,否则flase,2.当p的串大于1且p[1]=='*"时, 需要匹配'*',方法有当前s匹配p[2],或者当前元素相等匹配s[1]和p,3.在以上条件不满足,也就是要么p==1或p[1]!='*',则只需要匹配当前的s和p
class Solution {
public:
bool isMatch(string s, string p) {
if (p.empty()) return s.empty();
if (p.size() > 1 && p[1] == '*')
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));
}
};