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。

 1 class Solution {

 2 public:

 3     bool wordBreak(string s, unordered_set<string> &dict) {

 4         int n = s.length();

 5         if (n == 0) return true;

 6         vector<bool> possible(n, false);

 7         

 8         for (int i = 0; i < n; ++i) {

 9             for (int j = i; j >= 0; --j) {

10                 if ((j == 0 || possible[j - 1]) && dict.find(s.substr(j, i - j + 1)) != dict.end()) {

11                     possible[i] = true;

12                     break;

13                 }

14             }

15         }

16         

17         return possible[n - 1];

18     }

19 };

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 < 0) {

            ret.push_back(str.substr(0, str.length() - 1));

            return;

        }

        

        for (int i = index; i >= 0; --i) {

            string tmp = s.substr(i, index - i + 1);

            if (dict.find(tmp) != dict.end()) {

                bt(s, dict, tmp + " " + str, i - 1, ret);

            }

        }

    }

};