[LeetCode]10. Regular Expression Matching

https://leetcode.com/problems/regular-expression-matching/

给的case：ab -- .* --> true 是因为*代表前一位重复零次或多次，前一位是. 所以可以转化成..

DP问题，状态转移方程如下：

1, If p.charAt(j) == s.charAt(i) : dp[i][j] = dp[i-1][j-1];
2, If p.charAt(j) == '.' : dp[i][j] = dp[i-1][j-1];
3, If p.charAt(j) == '*':
here are two sub conditions:
1 if p.charAt(j-1) != s.charAt(i) : dp[i][j] = dp[i][j-2] //in this case, a* only counts as empty
2 if p.charAt(i-1) == s.charAt(i) or p.charAt(i-1) == '.':
dp[i][j] = dp[i-1][j] //in this case, a* counts as multiple a
or dp[i][j] = dp[i][j-1] // in this case, a* counts as single a
or dp[i][j] = dp[i][j-2] // in this case, a* counts as empty

初始化DP数组时是s为空，如果当前p位置i为*，同时i - 2为true，此时可以使得*代表零个前一位置数，从而当前位置为true

public class Solution {    public boolean isMatch(String s, String p) {        boolean[][] dp = new boolean[s.length() + 1][p.length() + 1];        dp[0][0] = true;        for (int i = 0; i < p.length(); i++) {            if (p.charAt(i) == '*' && dp[0][i - 1]) {                dp[0][i + 1] = true;            }        }        for (int i = 0; i < s.length(); i++) {            for (int j = 0; j < p.length(); j++) {                if (p.charAt(j) == '.') {                    dp[i + 1][j + 1] = dp[i][j];                }                if (p.charAt(j) == s.charAt(i)) {                    dp[i + 1][j + 1] = dp[i][j];                }                if (p.charAt(j) == '*') {                    if (p.charAt(j - 1) != s.charAt(i)) {                        dp[i + 1][j + 1] = dp[i + 1][j - 1];                    }                    if (p.charAt(j - 1) == s.charAt(i) | p.charAt(j - 1) == '.') {                        dp[i + 1][j + 1] = (dp[i + 1][j] || dp[i + 1][j - 1] || dp[i][j + 1]);                    }                }            }        }        return dp[s.length()][p.length()];    }}