Leetcode | Work Break I & II

Work Break I

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".

DP很容易就出来了。possible[i]保存[0,i]是否可以被分割的结果。

possible[i] = true, 当存在possible[k] = true，且[k,i]是dict里的一个word时。否则possible[i] = false。

这种是自底而下的。

 class Solution {
 public:
     bool wordBreak(string s, unordered_set<string> &dict) {
         int n = s.length();
         if (n == ) return true;
         vector<bool> possible(n, false);
 
         for (int i = ; i < n; ++i) {
             for (int j = i; j >= ; --j) {
                 if ((j ==  || possible[j - ]) && dict.find(s.substr(j, i - j + )) != dict.end()) {
                     possible[i] = true;
                     break;
                 }
             }
         }
 
         return possible[n - ];
     }
 };

算法的时间复杂度最坏情况是O(n^2)，空间复杂度是O(n)。

网上也有人用前缀树（Trie树、字典树）实现的。私以为用前缀树还得先将dict里的所有word插进去，时间复杂度为O(n*l+s)，l为word的最大长度，s为dict的大小。如果dict的大小比n大得多，那么整个开销也是不菲的。

只要稍微将上面的代码优化一下，先求出word的最大长度，那么时间复杂度也可以优化到O(n*l+s)。

Work Break II

Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word.

Return all such possible sentences.

For example, given
s = "catsanddog",
dict = ["cat", "cats", "and", "sand", "dog"].

A solution is ["cats and dog", "cat sand dog"].

直接用回溯就可以了。自顶而下。

class Solution {
public:
    vector<string> wordBreak(string s, unordered_set<string> &dict) {
        vector<string> ret;
        bt(s, dict, "", s.length(), ret);
        return ret;
    }
 
    void bt(string &s, unordered_set<string> &dict, string str, int index, vector<string> &ret) {
        if (index < ) {
            ret.push_back(str.substr(, str.length() - ));
            return;
        }
 
        for (int i = index; i >= ; --i) {
            string tmp = s.substr(i, index - i + );
            if (dict.find(tmp) != dict.end()) {
                bt(s, dict, tmp + " " + str, i - , ret);
            }
        }
    }
};