LeetCode - 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".

 

Solution1:

递归调用。

private Map<String, Boolean> map = new HashMap<String, Boolean>();
public boolean wordBreak(String s, Set<String> dict) {
    if(dict.contains(s)) return true;
    if(map.containsKey(s)) {
        return map.get(s);
    } 
    for(int i=0; i<s.length(); i++) {
        String prefix = s.substring(0, i+1);
        if(dict.contains(prefix)) {
            String suffix = s.substring(i+1);
            if(wordBreak(suffix, dict)) {
                return true;
            }
        }
    }
    map.put(s, false);
    return false;
}

Solution2:

动态规划。

f[i] = true  if  s[0,i]在dict里面

     = true  if  f[k] == true 并且s[k+1,j]在dict里面，0<k<i

     = false if  no such k exist.

public boolean wordBreak(String s, Set<String> dict) {
    s = "#"+s;
    boolean[] f = new boolean[s.length()];
    f[0] = true;
    for(int i=1; i<s.length(); i++) {
        for(int j=0; j<i; j++) {
            f[i] = f[j] && dict.contains(s.substring(j+1, i+1));
            if(f[i]) break;
        }
    }
    return f[s.length()-1];
 }

 

Reference:

http://fisherlei.blogspot.com/2013/11/leetcode-word-break-solution.html