• 10. Regular Expression Matching


    Given an input string (s) and a pattern (p), 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).

    Note:

    • s could be empty and contains only lowercase letters a-z.
    • p could be empty and contains only lowercase letters a-z, and characters like . or *.

    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
    
    
    class Solution {
        public boolean isMatch(String text, String pattern) {
    
            //递归结束条件
            if(pattern.isEmpty()) return text.isEmpty();
            //判断 text 是否为空,防止越界,如果 text 为空, 表达式直接判为 false, text.charAt(0)就不会执行了
            boolean firstmatch = (!text.isEmpty() && (pattern.charAt(0) == text.charAt(0) || pattern.charAt(0) == '.'));
            //两种情况
            //pattern 直接跳过两个字符。表示 * 前边的字符出现 0 次
            //pattern 不变,例如 text = aa ,pattern = a*,第一个 a 匹配,然后 text 的第二个 a 接着和 pattern 的第一个 a 进行匹配。表示 * 用前一个字符替代。
            if(pattern.length() >= 2 && pattern.charAt(1) == '*'){
                return (isMatch(text, pattern.substring(2)) || firstmatch && isMatch(text.substring(1), pattern));
            }
            else    return firstmatch && isMatch(text.substring(1), pattern.substring(1));
        }
    }
    
    
    
     

    https://leetcode.windliang.cc/leetCode-10-Regular-Expression-Matching.html

    厉害了 我的哥

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  • 原文地址:https://www.cnblogs.com/wentiliangkaihua/p/10353662.html
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