Word Break

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".

 方法跟 Palindrome Partitioning II 基本一样

possible [ i ] 用来记录前 i 个元素 是不是都是dict的。

isWord[ i ][ j ] 用来记录 string 的 从 i 到 j 位置是不是在 dict 里面

里面一点可以优化的地方 是求出 dict 里面最长单词的长度，用来作为边界条件。

public class Solution {
    public boolean wordBreak(String s, Set<String> dict) {
        if (s == null || s.length() ==0) {
            return false;
        }
        int l = s.length();
		boolean[] possible = new boolean[l + 1];
		possible[0] = true;
		int maxLength = longest(dict);
		boolean[][] inDict = isWord(s, maxLength, dict);
		for (int i = 0; i < l; i++) {
		    for(int j = 0; j <=maxLength; j++) {
		        if (j > i ) {
		            break;
		        }
		        if (possible[i - j] &&inDict[i - j][i]) {
		            possible[i+1] = true;
		            break;
		        }
		    }
		}
		return possible[l];
		
    }
    private boolean[][] isWord(String s, int maxLength, Set<String> dict) {
        boolean[][] res = new boolean[s.length()][s.length()];
        for (int i = 0; i < s.length(); i++) {
			for (int j = 1; j <= maxLength; j++) {
			    if (i + j > s.length()){
			        break;
			    }
				res[i][i+j-1] = dict.contains(s.substring(i,i+j ));
			}
		}
		return res;
    }
    private int longest (Set<String> dict) {
        int res = 0;
        for (String word: dict) {
            res = Math.max(res,word.length());
        }
        return res;
    }
}

第二次写：

public class Solution {
    public boolean wordBreak(String s, Set<String> dict) {
        if (dict == null || dict.size() == 0) {
            return false;
        }
        boolean[] result = new boolean[s.length() + 1];
        result[0] = true;
        for (int i = 0; i < s.length(); i++) {
            for (int j = 0; j <= i; j++) {
                if (result[j] && dict.contains(s.substring(j, i + 1))) {
                    result[i + 1] = true;
                    break;
                }
            }
        }
        return result[s.length()];
    }
}