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LeetCode-139-Word-Break.java
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LeetCode-139-Word-Break.java
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/*
LeetCode: https://leetcode.com/problems/word-break/
LintCode: http://www.lintcode.com/problem/word-break/
JiuZhang: http://www.jiuzhang.com/solutions/word-break/
ProgramCreek: http://www.programcreek.com/2012/12/leetcode-solution-word-break/
Analysis:
1.DFS
Time Limit Exceed
Time O(N^2*wordDict Size) ???
2.DP
Time O(N*dict size)
*/
public class Solution {
// 2.DP
public boolean wordBreak(String s, Set<String> wordDict) {
// state
boolean[] state = new boolean[s.length() + 1];
state[0] = true; //why initialized true? state[0] has function of dummy node in List problem
// canculate state
for(int i = 0; i <= s.length(); i++){
if(!state[i]) continue; //if start char is not true, don't need to continue calculating
for(String word : wordDict){
int len = word.length();
int end = i + len; //notice, i is start, end is the next char index of next word
if(end > s.length()) continue;
if(state[end]) continue; //optimize: if has already true, we don't need to calculate
if(s.substring(i, end).equals(word)){
state[end] = true;
}
}
}
return state[s.length()];
}
// 1.DFS
// Time Limit Exceed
// public boolean wordBreak(String s, Set<String> wordDict) {
// return wordBreak(s, wordDict, 0);
// }
// private boolean wordBreak(String s, Set<String> wordDict, int start){
// // end condition
// if(start == s.length()) return true;
// for(int i = start; i < s.length(); i++){
// for(String str : wordDict){
// int end = i + str.length(); //Corner case: think carefully the end
// if(end > s.length()) continue;
// if(s.substring(i, end).equals(str)){
// if(wordBreak(s, wordDict, end)) return true;
// }
// }
// }
// return false;
// }
}
// // 1. Build Trie + DFS
// class Solution {
// private class Node {
// char c;
// HashMap<Character, Node> children = new HashMap<>();
// String word = null;
// public Node (char c) {
// this.c = c;
// }
// }
// private class Trie {
// Node root;
// public Trie () {
// root = new Node('#');
// }
// public void add(String s) {
// Node curr = root;
// for (char c : s.toCharArray()) {
// if (curr.children.containsKey(c)) {
// curr = curr.children.get(c);
// } else {
// Node newNode = new Node(c);
// curr.children.put(c, newNode);
// curr = newNode;
// }
// }
// curr.word = s;
// }
// public Node getRoot() {
// return root;
// }
// }
// public boolean wordBreak(String s, List<String> wordDict) {
// Trie trie = new Trie();
// for (String str : wordDict) {
// trie.add(str);
// }
// return recursive(s, trie.getRoot());
// }
// private boolean recursive(String s, Node root) {
// if (s.length() == 0) return true;
// Node curr = root;
// for (char c : s.toCharArray()) {
// if (!curr.children.containsKey(c)) return false;
// curr = curr.children.get(c);
// if (curr.word != null) {
// if (recursive(s.substring(curr.word.length()), root)){
// return true;
// }
// }
// }
// return false;
// }
// }
// 2. Build Trie + Memorized DFS
/*
https://leetcode.com/problems/word-break/discuss/181632/Java-Solution-%3A-Trie-%2B-memoization
*/
// class Solution {
// private class Node {
// char c;
// HashMap<Character, Node> children = new HashMap<>();
// String word = null;
// public Node (char c) {
// this.c = c;
// }
// }
// private class Trie {
// Node root;
// public Trie () {
// root = new Node('#');
// }
// public void add(String s) {
// Node curr = root;
// for (char c : s.toCharArray()) {
// if (curr.children.containsKey(c)) {
// curr = curr.children.get(c);
// } else {
// Node newNode = new Node(c);
// curr.children.put(c, newNode);
// curr = newNode;
// }
// }
// curr.word = s;
// }
// public Node getRoot() {
// return root;
// }
// }
// public boolean wordBreak(String s, List<String> wordDict) {
// Trie trie = new Trie();
// for (String str : wordDict) {
// trie.add(str);
// }
// HashMap<String, Boolean> map = new HashMap<>();
// map.put("", true);
// return recursive(s, trie.getRoot(), map);
// }
// private boolean recursive(String s, Node root, HashMap<String, Boolean> map) {
// if (map.containsKey(s)) return map.get(s);
// Node curr = root;
// for (char c : s.toCharArray()) {
// if (!curr.children.containsKey(c)) return false;
// curr = curr.children.get(c);
// if (curr.word != null) {
// String next = s.substring(curr.word.length());
// if (recursive(next, root, map)) {
// map.put(next, true);
// return true;
// } else {
// map.put(next, false);
// }
// }
// }
// return false;
// }
// }
// 3. DP (pull)
/*
subproblem:
dp[i] - if the s.substring(0, i + 1) could be break into words in wordDict
recurrence relation:
dp[i] |= dp[i - w.length()], w belongs to wordDict
init:
dp[0] = true; // empty string
ans:
dp[n]
*/
// class Solution {
// public boolean wordBreak(String s, List<String> wordDict) {
// int n = s.length();
// // subproblem
// boolean[] dp = new boolean[n + 1];
// // init
// dp[0] = true;
// // recurrence relation
// for (int i = 1; i <= n; i++) {
// for (String w : wordDict) {
// if (i < w.length()) continue;
// if (w.equals(s.substring(i - w.length(), i))) {
// if (dp[i - w.length()]) {
// dp[i] = true;
// continue;
// }
// }
// }
// }
// // ans
// return dp[n];
// }
// }
// 4. DP (push)
/*
subproblem:
dp[i] - if the s.substring(0, i + 1) could be break into words in wordDict
recurrence relation:
dp[i] |= dp[i - w.length()], w belongs to wordDict
init:
dp[0] = true; // empty string
ans:
dp[n]
*/
class Solution {
public boolean wordBreak(String s, List<String> wordDict) {
int n = s.length();
// subproblem
boolean[] dp = new boolean[n + 1];
// init
dp[0] = true;
// recurrence relation
for (int i = 0; i <= n; i++) {
for (String w : wordDict) {
if (dp[i] == false) continue;
if (i + w.length() > n) continue;
if (w.equals(s.substring(i, i + w.length()))) {
dp[i + w.length()] = true;
if (i + w.length() == n) return true;
}
}
}
// ans
return dp[n];
}
}