Word Break(Java)

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”.

1. Naive Approach

This problem can be solve by using a naive approach, which is trivial. A discussion can always start from that though.

public class Solution {
public boolean wordBreak(String s, Set dict) {
return wordBreakHelper(s, dict, 0);
}

public boolean wordBreakHelper(String s, Set<String> dict, int start){    if(start == s.length())         return true;    for(String a: dict){        int len = a.length();        int end = start+len;        //end index should be <= string length        if(end > s.length())             continue;        if(s.substring(start, start+len).equals(a))            if(wordBreakHelper(s, dict, start+len))                return true;    }    return false;}

}
Time is O(n^2) and exceeds the time limit.

1. Dynamic Programming

The key to solve this problem by using dynamic programming approach:

Define an array t[] such that t[i]==true => 0-(i-1) can be segmented using dictionary
Initial state t[0] == true
public class Solution {
public boolean wordBreak(String s, Set dict) {
boolean[] t = new boolean[s.length()+1];
t[0] = true; //set first to be true, why?
//Because we need initial state

    for(int i=0; i<s.length(); i++){        //should continue from match position        if(!t[i])             continue;        for(String a: dict){            int len = a.length();            int end = i + len;            if(end > s.length())                continue;            if(t[end]) continue;            if(s.substring(i, end).equals(a)){                t[end] = true;            }        }    }    return t[s.length()];}

}
Time: O(string length * dict size).

1. Java Solution 3 - Simple and Efficient

In Solution 2, if the size of the dictionary is very large, the time is bad. Instead we can solve the problem in O(n^2) time (n is the length of the string).

public boolean wordBreak(String s, Set wordDict) {
int[] pos = new int[s.length()+1];

Arrays.fill(pos, -1);pos[0]=0;for(int i=0; i<s.length(); i++){    if(pos[i]!=-1){        for(int j=i+1; j<=s.length(); j++){            String sub = s.substring(i, j);            if(wordDict.contains(sub)){                pos[j]=i;            }        }     }}return pos[s.length()]!=-1;

}
转载地址：http://www.programcreek.com/2012/12/leetcode-solution-word-break/