[leetcode]10. Regular Expression Matching(Java)

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

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).The function prototype should be:bool isMatch(const char *s, const char *p)Some examples:isMatch("aa","a") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "a*") → trueisMatch("aa", ".*") → trueisMatch("ab", ".*") → trueisMatch("aab", "c*a*b") → true

package go.jacob.day707;/*  *  * 动态规划DP的应用  * Runtime: 29 ms.Your runtime beats 81.88 % of java submissions. */public class Demo1 {/* * 如果对动态规划不太了解，可以参考博客：http://blog.csdn.net/zjkc050818/article/details/74532023 * 以及博客中的视频。 */public boolean isMatch(String s, String p) {if (s == null || p == null)return false;int m = s.length(), n = p.length();boolean[][] res = new boolean[m + 1][n + 1];res[0][0] = true;//s.charAt(i)与p.charAt(j)的结果是存储在res[i+1][j+1]中//所以res[0][i]其实是存储p中与0个字符(s.charAt(-1)不存在)的匹配结果for (int i = 0; i < n; i++) {if (p.charAt(i) == '*' && res[0][i - 1])res[0][i + 1] = true;}for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {if (p.charAt(j) == '.')res[i + 1][j + 1] = res[i][j];if (p.charAt(j) == s.charAt(i))res[i + 1][j + 1] = res[i][j];if (p.charAt(j) == '*') {if (s.charAt(i) != p.charAt(j - 1) && p.charAt(j - 1) != '.')res[i + 1][j + 1] = res[i + 1][j - 1];else {//res[i + 1][j - 1] 表示*一个都不匹配;//res[i + 1][j]表示匹配一个 //res[i][j + 1]表示匹配多个res[i + 1][j + 1] = res[i + 1][j - 1] || res[i + 1][j] || res[i][j + 1];}}}}return res[m][n];}}