leetcode/10.regular-expression-matching.java at master · ShawnLi1014/leetcode · GitHub

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/*
 * @lc app=leetcode id=10 lang=java
 *
 * [10] Regular Expression Matching
 *
 * https://leetcode.com/problems/regular-expression-matching/description/
 *
 * algorithms
 * Hard (25.55%)
 * Likes:    2914
 * Dislikes: 553
 * Total Accepted:    331.1K
 * Total Submissions: 1.3M
 * Testcase Example:  '"aa"\n"a"'
 *
 * 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 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
 *
 *
 */

/*
    dp[i][j] = string s with length i matches string p with len j
*/
class Solution {
    public boolean isMatch(String s, String p) {
        if (s == null || p == null) {
            return false;
        }
        boolean[][] state = new boolean[s.length() + 1][p.length() + 1];
        state[0][0] = true;
        // Consider p.length() == 0, always false, already included
        // Consider s.length() == 0
        for (int j = 1; j < state[0].length; j++) {
            // Only current character is '*'
            if (p.charAt(j - 1) == '*' && state[0][j - 2]) {
                state[0][j] = true;
            }
        }

        for (int i = 1; i < state.length; ++i) {
            for (int j = 1; j < state[0].length; ++j) {
                if (s.charAt(i - 1) == p.charAt(j - 1) || p.charAt(j - 1) == '.') {
                    state[i][j] = state[i - 1][j - 1];
                } else if (p.charAt(j - 1) == '*') {
                    if (s.charAt(i - 1) != p.charAt(j - 2) && p.charAt(j - 2) != '.') {
                        state[i][j] = state[i][j - 2];
                    } else {
                        state[i][j] = state[i - 1][j] || state[i][j - 1] || state[i][j - 2];
                    }
                }
            }
        }
        return state[s.length()][p.length()];
    }
}