algorithm-class/大厂刷题班/class31/Problem_0139_WordBreak.java

package class31;

import java.util.List;

// lintcode也有测试，数据量比leetcode大很多 : https://www.lintcode.com/problem/107/
public class Problem_0139_WordBreak {

	public static class Node {
		public boolean end;
		public Node[] nexts;

		public Node() {
			end = false;
			nexts = new Node[26];
		}
	}

	public static boolean wordBreak1(String s, List<String> wordDict) {
		Node root = new Node();
		for (String str : wordDict) {
			char[] chs = str.toCharArray();
			Node node = root;
			int index = 0;
			for (int i = 0; i < chs.length; i++) {
				index = chs[i] - 'a';
				if (node.nexts[index] == null) {
					node.nexts[index] = new Node();
				}
				node = node.nexts[index];
			}
			node.end = true;
		}
		char[] str = s.toCharArray();
		int N = str.length;
		boolean[] dp = new boolean[N + 1];
		dp[N] = true; // dp[i]  word[i.....] 能不能被分解
		// dp[N] word[N...]  -> ""  能不能够被分解 
		// dp[i] ... dp[i+1....]
		for (int i = N - 1; i >= 0; i--) {
			// i
			// word[i....] 能不能够被分解
			// i..i    i+1....
			// i..i+1  i+2...
			Node cur = root;
			for (int end = i; end < N; end++) {
				cur = cur.nexts[str[end] - 'a'];
				if (cur == null) {
					break;
				}
				// 有路！
				if (cur.end) {
					// i...end 真的是一个有效的前缀串  end+1....  能不能被分解
					dp[i] |= dp[end + 1];
				}
				if (dp[i]) {
					break;
				}
			}
		}
		return dp[0];
	}

	public static int wordBreak2(String s, List<String> wordDict) {
		Node root = new Node();
		for (String str : wordDict) {
			char[] chs = str.toCharArray();
			Node node = root;
			int index = 0;
			for (int i = 0; i < chs.length; i++) {
				index = chs[i] - 'a';
				if (node.nexts[index] == null) {
					node.nexts[index] = new Node();
				}
				node = node.nexts[index];
			}
			node.end = true;
		}
		char[] str = s.toCharArray();
		int N = str.length;
		int[] dp = new int[N + 1];
		dp[N] = 1;
		for (int i = N - 1; i >= 0; i--) {
			Node cur = root;
			for (int end = i; end < N; end++) {
				cur = cur.nexts[str[end] - 'a'];
				if (cur == null) {
					break;
				}
				if (cur.end) {
					dp[i] += dp[end + 1];
				}
			}
		}
		return dp[0];
	}

}