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package class31;
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import java.util.ArrayList;
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import java.util.List;
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public class Problem_0140_WordBreakII {
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public static class Node {
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public String path;
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public boolean end;
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public Node[] nexts;
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public Node() {
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path = null;
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end = false;
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nexts = new Node[26];
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}
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}
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public static List<String> wordBreak(String s, List<String> wordDict) {
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char[] str = s.toCharArray();
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Node root = gettrie(wordDict);
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boolean[] dp = getdp(s, root);
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ArrayList<String> path = new ArrayList<>();
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List<String> ans = new ArrayList<>();
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process(str, 0, root, dp, path, ans);
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return ans;
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}
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// str[index.....] 是要搞定的字符串
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// dp[0...N-1] 0... 1.... 2... N-1... 在dp里
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// root 单词表所有单词生成的前缀树头节点
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// path str[0..index-1]做过决定了，做的决定放在path里
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public static void process(char[] str, int index, Node root, boolean[] dp, ArrayList<String> path,
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List<String> ans) {
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if (index == str.length) {
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StringBuilder builder = new StringBuilder();
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for (int i = 0; i < path.size() - 1; i++) {
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builder.append(path.get(i) + " ");
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}
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builder.append(path.get(path.size() - 1));
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ans.add(builder.toString());
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} else {
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Node cur = root;
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for (int end = index; end < str.length; end++) {
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// str[i..end] （能不能拆出来）
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int road = str[end] - 'a';
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if (cur.nexts[road] == null) {
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break;
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}
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cur = cur.nexts[road];
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if (cur.end && dp[end + 1]) {
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// [i...end] 前缀串
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// str.subString(i,end+1) [i..end]
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path.add(cur.path);
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process(str, end + 1, root, dp, path, ans);
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path.remove(path.size() - 1);
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}
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}
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}
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}
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public static Node gettrie(List<String> wordDict) {
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Node root = new Node();
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for (String str : wordDict) {
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char[] chs = str.toCharArray();
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Node node = root;
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int index = 0;
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for (int i = 0; i < chs.length; i++) {
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index = chs[i] - 'a';
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if (node.nexts[index] == null) {
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node.nexts[index] = new Node();
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}
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node = node.nexts[index];
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}
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node.path = str;
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node.end = true;
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}
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return root;
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}
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public static boolean[] getdp(String s, Node root) {
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char[] str = s.toCharArray();
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int N = str.length;
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boolean[] dp = new boolean[N + 1];
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dp[N] = true;
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for (int i = N - 1; i >= 0; i--) {
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Node cur = root;
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for (int end = i; end < N; end++) {
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int path = str[end] - 'a';
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if (cur.nexts[path] == null) {
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break;
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}
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cur = cur.nexts[path];
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if (cur.end && dp[end + 1]) {
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dp[i] = true;
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break;
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}
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}
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}
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return dp;
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}
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}
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