Word Break

Change-Id: Ib9573f25a92cb9f43bb03f7d304740199407c4d3

Author: applewjg

WordBreak.java

@@ -0,0 +1,33 @@
+/*
+ Author:     Andy, nkuwjg@gmail.com
+ Date:       Jan 26, 2015
+ Problem:    Word Break
+ Difficulty: Easy
+ Source:     http://oj.leetcode.com/problems/word-break/
+ Notes:
+ Given a string s and a dictionary of words dict, determine if s can be segmented into 
+ a space-separated sequence of one or more dictionary words.
+ For example, given
+ s = "leetcode",
+ dict = ["leet", "code"].
+ Return true because "leetcode" can be segmented as "leet code".
+
+ Solution: dp.
+*/
+
+public class Solution {
+    public boolean wordBreak(String s, Set<String> dict) {
+        int n = s.length();
+        boolean[] dp = new boolean[n+1];
+        dp[n] = true;
+        for (int i = n - 1; i >= 0; --i) {
+            for (int j = i; j < n; ++j) {
+                if (dict.contains(s.substring(i,j+1)) && dp[j+1]) {
+                    dp[i] = true;
+                    break;
+                }
+            }
+        }
+        return dp[0];
+    }
+}

WordBreakII.java

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+/*
+ Author:     Andy, nkuwjg@gmail.com
+ Date:       Jan 29, 2015
+ Problem:    Word Break II
+ Difficulty: Easy
+ Source:     http://oj.leetcode.com/problems/word-break-ii/
+ Notes:
+ 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"].
+
+ Solution: check before constructing the sentences.
+*/
+
+public class Solution {
+    public List<String> wordBreak(String s, Set<String> dict) {
+        List<String> res = new ArrayList<String>();
+        int n = s.length();
+        boolean[] canBreak = new boolean[n+1];
+        canBreak[n] = true;
+        for (int i = n - 1; i >= 0; --i) {
+            for (int j = i; j < n; ++j) {
+                if (dict.contains(s.substring(i,j+1)) && canBreak[j+1]) {
+                    canBreak[i] = true;
+                    break;
+                }
+            }
+        }
+        if (canBreak[0] == false) return res;
+        wordBreakRe(s, dict, "", 0, res);
+        return res;
+    }
+    public void wordBreakRe(String s, Set<String> dict, String path, int start, List<String> res) {
+        if (start == s.length()) {
+            res.add(path);
+            return;
+        }
+        if (path.length() != 0) path = path + " ";
+        for (int i = start; i < s.length(); ++i) {
+            String word = s.substring(start, i + 1);
+            if (dict.contains(word) == false) continue;
+            wordBreakRe(s, dict, path + word, i + 1, res);
+        }
+    }
+}