LeetCode 291. Word Pattern II（单词模式II）

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原题网址：https://leetcode.com/problems/word-pattern-ii/

Given a pattern and a string str, find if str follows the same pattern.

Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty substring in str.

Examples:

pattern = "abab", str = "redblueredblue" should return true.
pattern = "aaaa", str = "asdasdasdasd" should return true.
pattern = "aabb", str = "xyzabcxzyabc" should return false.

Notes:
You may assume both pattern and str contains only lowercase letters.

方法：深度优先搜索。

public class Solution {    private int pcount = 0;    private int[] acount = new int[26];    private int[] ntoa = new int[26];    private int[] aton = new int[26];    private char[] sa;    private String str;    private boolean matched;    private char[] pa;    private void find(int[] lens, int len, int step) {        if (matched) return;        if (len>sa.length) return;        else if (len==sa.length) {            if (step!=lens.length) return;            // len==sa.length && step==lens.length            int pos = 0;            String[] map = new String[26];            for(int i=0; i<pa.length; i++) {                String s = str.substring(pos, pos+lens[aton[pa[i]-'a']]);                if (map[aton[pa[i]-'a']] == null) map[aton[pa[i]-'a']] = s;                else if (!map[aton[pa[i]-'a']].equals(s)) return;                pos += lens[aton[pa[i]-'a']];            }            for(int i=0; i<pa.length-1; i++) {                for(int j=i+1; j<pa.length; j++) {                    if (pa[i] != pa[j] && map[aton[pa[i]-'a']].equals(map[aton[pa[j]-'a']])) return;                }            }            matched = true;            return;        } else if (step==lens.length) return;        for(int i=1; len + acount[this.ntoa[step]]*i <= sa.length; i ++) {            lens[step] = i;            find(lens, len + acount[this.ntoa[step]]*i, step + 1);        }    }            public boolean wordPatternMatch(String pattern, String str) {        this.str = str;        this.sa = str.toCharArray();        this.pa = pattern.toCharArray();        for(int i=0; i<this.pa.length; i++) {            int j=this.pa[i]-'a';            if (this.acount[j] == 0) {                this.aton[j] = this.pcount;                this.ntoa[this.pcount++] = j;            }            this.acount[j] ++;        }        find(new int[this.pcount], 0, 0);        return matched;    }}

优化：

public class Solution {    String str;    char[] pa;    int pcount;    int[] aton = new int[26];    int[] ntoa = new int[26];    int[] acount = new int[26];    private boolean match(int[] lens, int len, int step) {        if (step == lens.length) {            if (len != str.length()) return false;            String[] map = new String[26];            int pos = 0;            for(int i=0; i<pa.length; i++) {                String s = str.substring(pos, pos+lens[aton[pa[i]-'a']]);                if (map[aton[pa[i]-'a']] == null) map[aton[pa[i]-'a']] = s;                else if (!s.equals(map[aton[pa[i]-'a']])) return false;                pos += lens[aton[pa[i]-'a']];            }            for(int i=0; i<pcount-1; i++) {                for(int j=i+1; j<pcount; j++) {                    if (map[i].equals(map[j])) return false;                }            }            return true;        }        for(int i=1; len+acount[ntoa[step]]*i+(lens.length-1-step)<=str.length(); i++) {            lens[step] = i;            if (match(lens, len+acount[ntoa[step]]*i, step+1)) return true;        }        return false;    }    public boolean wordPatternMatch(String pattern, String str) {        pa = pattern.toCharArray();        for(int i=0; i<pa.length; i++) {            int j=pa[i]-'a';            if (acount[j]==0) {                aton[j]=pcount;                ntoa[pcount]=j;                pcount++;            }            acount[j]++;        }        this.str = str;        return match(new int[pcount], 0, 0);    }}

优化：

public class Solution {    private boolean find(char[] pattern, int step, int len, String str, String[] map, Map<String, Character> inverse) {        if (step == pattern.length) {            if (len == str.length()) return true;            return false;        }        if (map[pattern[step] - 'a'] != null) {            if (len + map[pattern[step] - 'a'].length() > str.length()) return false;            if (!map[pattern[step] - 'a'].equals(str.substring(len, len + map[pattern[step] - 'a'].length()))) return false;            return find(pattern, step + 1, len + map[pattern[step] - 'a'].length(), str, map, inverse);        }        int from = 1, to = (str.length() - len) - (pattern.length - 1 - step);        if (from > to) return false;        if (step == pattern.length - 1) from = to;        for(int i = from; i <= to; i++) {            String sub = str.substring(len, len + i);            Character ch = inverse.get(sub);            if (ch != null) continue;            inverse.put(sub, pattern[step]);            map[pattern[step] - 'a'] = sub;            if (find(pattern, step + 1, len + i, str, map, inverse)) return true;            inverse.remove(sub);            map[pattern[step] - 'a'] = null;        }        return false;    }    public boolean wordPatternMatch(String pattern, String str) {        return find(pattern.toCharArray(), 0, 0, str, new String[26], new HashMap<>());    }}

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