[leetcode-139]Word Break（java）

问题描述：

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”.

分析：这种问题已经出现了很多次了，本质上就是要求把大问题切割为小问题，虽然表面上切割为小问题再归纳为大问题好像多此一举，但实际上这样能省很多时间。

最开始使用递归的思想，结果TLE，最后改成了DP算法，顺利AC，

代码如下：304ms

[code]public class Solution {
  public boolean wordBreak(String s,Set<String> wordDict){
        int length = s.length();
        boolean[] words = new boolean[length];

        for(int i = 1;i<=length;i++){
            int j;
            for(j = 0;j<i;j++){
                String substr = s.substring(j,i);
                if(wordDict.contains(substr) && (j==0 || words[j-1])) {
                    words[i-1] = true;
                    break;
                }
            }
            if(j == i)
                words[i-1] = false;
        }
        return words[length-1];
    }
}

递归版本：TLE

[code]public boolean wordBreak(String s, Set<String> wordDict) {
        if(s.length()==0)
            return false;
        if(wordDict.contains(s))
            return true;
        for(int i = 1;i<=s.length();i++){
            String substring = s.substring(0,i);
            if(wordDict.contains(substring) && wordBreak(s.substring(i),wordDict))
                return true;
        }
        return false;
    }