字符串、字典匹配算法

package StringAndArray;import java.util.HashSet;import java.util.Set;import java.util.regex.Matcher;import java.util.regex.Pattern;/** * 一字符串s与一个包含多个字符串的set字典dict，查询set字典中部分字符串是否能拼接成s *  * @author fh * */public class Word_Break {    public static void main(String[] args) {        Set<String> dict = null;        String s=null;        s="programcreek";        dict = new HashSet<String>();        dict.add("programcree");        dict.add("program");        dict.add("creek");        boolean b=wordBreak(s, dict);        System.out.println(s+"在字典dict是否可查询："+b);        // Regrex();    }    /**     * The key to solve this problem by using dynamic programming approach:     *      * Define an array t[] such that t[i]==true => 0-(i-1) can be segmented     * using dictionary Initial state t[0] == true     *      * Time: O(string length * dict size)     *      * One tricky part of this solution is the case:     *      * INPUT: "programcreek", ["programcree","program","creek"].     */    public static boolean wordBreak(String s, Set<String> dict) {        boolean[] t = new boolean[s.length() + 1];        t[0] = true;// 需要一个初始值        for (int i = 0; i < s.length(); i++) {            if (!t[i])// 该字符与前面拼接的字符串在dict中不存在，则跳出本次循环                continue;            // 循环遍历字典，查出以i为起点的所有匹配子字符串            for (String sub : dict) {                int len = sub.length();                int end = i + len;                //剩余待匹配字符串长度小于字典中的匹配子字符串                if (end > s.length())                    continue;                if(s.substring(i, end).equals(sub)){                    t[end]=true;                    if(end==s.length()){                        return true;                    }                }            }        }        return false;    }    // 正则法解题    private static void Regrex() {        HashSet<String> dict = new HashSet<String>();        dict.add("go");        dict.add("goal");        dict.add("goals");        dict.add("special");        StringBuilder sb = new StringBuilder();        for (String s : dict) {            sb.append(s + "|");        }        String pattern = sb.toString().substring(0, sb.length() - 1);        pattern = "(" + pattern + ")*";        System.err.println(pattern);        Pattern p = Pattern.compile(pattern);        Matcher m = p.matcher("goalgoalspecial");        if (m.matches()) {            System.out.println("match");        }    }}