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Commit2c41774

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author
applewjg
committed
Word Break
Change-Id: Ib9573f25a92cb9f43bb03f7d304740199407c4d3
1 parentc4271e5 commit2c41774

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2 files changed

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2 files changed

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‎WordBreak.java

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

‎WordBreakII.java

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/*
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Author: Andy, nkuwjg@gmail.com
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Date: Jan 29, 2015
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Problem: Word Break II
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Difficulty: Easy
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Source: http://oj.leetcode.com/problems/word-break-ii/
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Notes:
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Given a string s and a dictionary of words dict, add spaces in s to
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construct a sentence where each word is a valid dictionary word.
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Return all such possible sentences.
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For example, given
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s = "catsanddog",
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dict = ["cat", "cats", "and", "sand", "dog"].
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A solution is ["cats and dog", "cat sand dog"].
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Solution: check before constructing the sentences.
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*/
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publicclassSolution {
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publicList<String>wordBreak(Strings,Set<String>dict) {
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List<String>res =newArrayList<String>();
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intn =s.length();
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boolean[]canBreak =newboolean[n+1];
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canBreak[n] =true;
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for (inti =n -1;i >=0; --i) {
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for (intj =i;j <n; ++j) {
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if (dict.contains(s.substring(i,j+1)) &&canBreak[j+1]) {
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canBreak[i] =true;
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break;
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}
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}
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}
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if (canBreak[0] ==false)returnres;
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wordBreakRe(s,dict,"",0,res);
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returnres;
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}
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publicvoidwordBreakRe(Strings,Set<String>dict,Stringpath,intstart,List<String>res) {
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if (start ==s.length()) {
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res.add(path);
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return;
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}
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if (path.length() !=0)path =path +" ";
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for (inti =start;i <s.length(); ++i) {
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Stringword =s.substring(start,i +1);
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if (dict.contains(word) ==false)continue;
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wordBreakRe(s,dict,path +word,i +1,res);
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}
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}
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}

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