10. Regular Expression Matching

[Tcdian]

10. Regular Expression Matching

给你一个字符串 s 和一个字符规律 p，请你来实现一个支持 '.' 和 '*' 的正则表达式匹配。

'.' 匹配任意单个字符
'*' 匹配零个或多个前面的那一个元素

所谓匹配，是要涵盖 整个 字符串 s的，而不是部分字符串。

Note

• 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 *。

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 preceding 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

[Tcdian]

Solution

• JavaScript Solution

/**
 * @param {string} s
 * @param {string} p
 * @return {boolean}
 */
var isMatch = function(s, p) {
    const dp = Array.from(new Array(p.length + 1), () => new Array(s.length + 1));
    for(let i = 0; i <= p.length; i++) {
        for (let j = 0; j <= s.length; j++) {
            if (i === 0 && j === 0) {
                dp[i][j] = true;
            } else if (i == 0) {
                dp[i][j] = false;
            } else if (p[i - 1] === '.' || p[i - 1] === s[j - 1]) {
                dp[i][j] = dp[i - 1][j - 1] || false;
            } else if (p[i - 1] === '*') {
                if (dp[i - 2][j]) {
                    dp[i][j] = true;
                    continue;
                }
                if (p[i - 2] === '.' || p[i - 2] === s[j - 1]) {
                    dp[i][j] = dp[i][j - 1] || dp[i - 1][j];
                    continue;
                }
                dp[i][j] = false;
            } else {
                dp[i][j] = false;
            }
        }
    }
    return dp[p.length][s.length];
};

• TypeScript Solution

function isMatch(s: string, p: string): boolean {
    const dp: boolean[][] = Array.from(new Array(p.length + 1), () => new Array(s.length + 1));
    for(let i = 0; i <= p.length; i++) {
        for (let j = 0; j <= s.length; j++) {
            if (i === 0 && j === 0) {
                dp[i][j] = true;
            } else if (i == 0) {
                dp[i][j] = false;
            } else if (p[i - 1] === '.' || p[i - 1] === s[j - 1]) {
                dp[i][j] = dp[i - 1][j - 1] || false;
            } else if (p[i - 1] === '*') {
                if (dp[i - 2][j]) {
                    dp[i][j] = true;
                    continue;
                }
                if (p[i - 2] === '.' || p[i - 2] === s[j - 1]) {
                    dp[i][j] = dp[i][j - 1] || dp[i - 1][j];
                    continue;
                }
                dp[i][j] = false;
            } else {
                dp[i][j] = false;
            }
        }
    }
    return dp[p.length][s.length];
};