[leetcode] Word Break II

https://leetcode.com/problems/word-break-ii/

Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word.

Return all such possible sentences.

For example, given
s = "catsanddog",
dict = ["cat", "cats", "and", "sand", "dog"].

A solution is ["cats and dog", "cat sand dog"].

public class Solution {    private String s;    private Set<String> dict;    private int len;    private LinkedList<String> wds;    private List<String> sentences;    private int[][] idxes;// idxes[i][j] means i can reach j    boolean hasAns = false;    public List<String> wordBreak(String s, Set<String> wordDict) {        sentences = new ArrayList<String>();        if(null == s || "".equals(s)) {            return sentences;        }                init(s, wordDict);        checkReachable0();        if(hasAns) {        find0();        }        return sentences;    }        private void find0() {    for(int i=0, n=idxes[0][len]; i<=n; ++i) {    int k = idxes[0][i]+1;    wds.add(s.substring(0, k));        findRest(k);        wds.removeLast();    }}private void findRest(int st) {if(st == len) {addSentence();return;}for(int i=0, n=idxes[st][len]; i<=n; ++i) {int k = idxes[st][i] + 1;wds.add(s.substring(st, k));        findRest(k);        wds.removeLast();}}private void checkReachable0() {        for(int i=0; i<len; ++i) {            if(dict.contains(s.substring(0, i+1))) {            idxes[0][++idxes[0][len]] = i;                hasAns |= (i+1 == len);                checkReachable(i+1);            }        }}private void checkReachable(int st) {if(st < len && 0 > idxes[st][len]) {for(int i=st; i<len; ++i) {if(dict.contains(s.substring(st, i+1))) {idxes[st][++idxes[st][len]] = i;                hasAns |= (i+1 == len);                checkReachable(i+1);}}}}    private void addSentence() {Iterator<String> it= wds.iterator();StringBuilder sb = new StringBuilder();if(it.hasNext()) {sb.append(it.next());}while(it.hasNext()) {sb.append(" ").append(it.next());}sentences.add(sb.toString());}private void init(String s, Set<String> wordDict) {        this.len = s.length();        this.s = s;        this.dict = wordDict;        this.wds = new LinkedList<String>();        this.idxes = new int[len][len+1];        for(int i=0; i<len; ++i) {        // -1 means this idx can reach nowhere        idxes[i][len] = -1;        }    }}

public class Solution {private String s;private Set<String> dict;private int len;private LinkedList<String> wds;private List<String> sentences;// idxes[i][k]=j: means that i can reach j,k is within[0, count[i]]private int[][] idxes;// count[i]: the number of word which starts at i and exits in wordDictprivate int[] count;boolean hasAns = false;public List<String> wordBreak(String s, Set<String> wordDict) {sentences = new ArrayList<String>();if (null == s || "".equals(s)) {return sentences;}init(s, wordDict);checkReachable();if (hasAns) {find(0);}return sentences;}private void find(int st) {if (st == len) {addSentence();return;}for (int i = 0; i < count[st]; ++i) {int k = idxes[st][i] + 1;wds.add(s.substring(st, k));find(k);wds.removeLast();}}/** * 从0开始找，找到所有可能的路径. */private void checkReachable() {for (int i = 0; i < len; ++i) {if (dict.contains(s.substring(0, i + 1))) {idxes[0][count[0]++] = i;hasAns |= (i + 1 == len);checkReachableRest(i + 1);}}}private void checkReachableRest(int st) {if (st < len && 0 == count[st]) {// 遍历过一次的，就全部找到了;所以要有這個條件開著，否則會多次遍歷。for (int i = st; i < len; ++i) {if (dict.contains(s.substring(st, i + 1))) {idxes[st][count[st]++] = i;hasAns |= (i + 1 == len);checkReachableRest(i + 1);}}}}private void addSentence() {Iterator<String> it = wds.iterator();StringBuilder sb = new StringBuilder();if (it.hasNext()) {sb.append(it.next());}while (it.hasNext()) {sb.append(" ").append(it.next());}sentences.add(sb.toString());}private void init(String s, Set<String> wordDict) {this.len = s.length();this.s = s;this.dict = wordDict;this.wds = new LinkedList<String>();this.idxes = new int[len][len];this.count = new int[len];}}

public class Solution {    private String s;private Set<String> dict;private int len;private LinkedList<String> wds;private List<String> sentences;// idxes[i][k]=j: means that i can reach j,k is within[0, count[i]]private int[][] idxes;// count[i]: the number of word which starts at i and exits in wordDictprivate int[] count;boolean hasAns = false;public List<String> wordBreak(String s, Set<String> wordDict) {sentences = new ArrayList<String>();if (null == s || "".equals(s)) {return sentences;}init(s, wordDict);checkReachable();if (hasAns) {find(0);}return sentences;}private void find(int st) {if (st == len) {addSentence();return;}for (int i = 0; i < count[st]; ++i) {int k = idxes[st][i] + 1;wds.add(s.substring(st, k));find(k);wds.removeLast();}}/** * 从0开始找，找到所有可能的路径. */private void checkReachable() {Queue<Integer> q = new LinkedList<Integer>();q.add(0);while (!q.isEmpty()) {int st = q.poll();hasAns |= (st == len);if (st < len && count[st] == 0) {// 没有计算过st的countfor (int i = st; i < len; ++i) {if (dict.contains(s.substring(st, i + 1))) {q.add(i + 1);idxes[st][count[st]++] = i;}}}}}private void addSentence() {Iterator<String> it = wds.iterator();StringBuilder sb = new StringBuilder();if (it.hasNext()) {sb.append(it.next());}while (it.hasNext()) {sb.append(" ").append(it.next());}sentences.add(sb.toString());}private void init(String s, Set<String> wordDict) {this.len = s.length();this.s = s;this.dict = wordDict;this.wds = new LinkedList<String>();this.idxes = new int[len][len];this.count = new int[len];}}