9 Palindrome Number

Hard

Given an input string (s) and a pattern (p), implement regular expression matching with support for '.' and '*'.

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'.' 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:

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Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

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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:

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Input:
s = "ab"
p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".

Example 4:

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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:

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Input:
s = "mississippi"
p = "mis*is*p*."
Output: false

给定一个字符串 (s) 和一个字符模式 (p)。实现支持 '.' 和 '*' 的正则表达式匹配。

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'.' 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 (s) ，而不是部分字符串。

说明:

  • s 可能为空，且只包含从 a-z 的小写字母。
  • p 可能为空，且只包含从 a-z 的小写字母，以及字符 . 和 *。

示例 1:

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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:

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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:

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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:

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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:

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输入:
s = "mississippi"
p = "mis*is*p*."
输出: false

想法

解

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class Solution {
  public:
    bool isMatch(string s, string p)
    {
        if (p.size() <= 0) {
            return s.size() <= 0;
        }

        bool dp[s.size() + 1][p.size() + 1];
        memset(dp, 0, sizeof(bool) * (s.size() + 1) * (p.size() + 1));
        
        // Initialization
        dp[0][0] = true;
        // Check the situation "#*#*#*#*#*#*#*"
        for (int i = 0; i < p.size(); i++) {
            if (p[i] == '*' && dp[0][i - 1]) {
                dp[0][i + 1] = true;
            }
        }

        // dp[0][*] has been used, so dp[i+1][j+1] denotes s[i] <==> p[j]
        for (int i = 0; i < s.size(); i++) {
            for (int j = 0; j < p.size(); j++) {
                if (s[i] == p[j] || p[j] == '.') {
                    dp[i + 1][j + 1] = dp[i][j];
                }
                else if (p[j] == '*') {
                    if (s[i] == p[j - 1] || p[j - 1] == '.') {
                        dp[i + 1][j + 1] = dp[i][j + 1] || dp[i + 1][j] || dp[i + 1][j - 1];
                    }
                    else {
                        dp[i + 1][j + 1] = dp[i + 1][j - 1];
                    }
                }
            }
        }

        return dp[s.size()][p.size()];
    }
};

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