力扣算法——140WordBreakII【H】

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences.

Note:

• The same word in the dictionary may be reused multiple times in the segmentation.
• You may assume the dictionary does not contain duplicate words.

Example 1:

Input:
s = "catsanddog"
wordDict = ["cat", "cats", "and", "sand", "dog"]
Output:
[
  "cats and dog",
  "cat sand dog"
]

Example 2:

Input:
s = "pineapplepenapple"
wordDict = ["apple", "pen", "applepen", "pine", "pineapple"]
Output:
[
  "pine apple pen apple",
  "pineapple pen apple",
  "pine applepen apple"
]
Explanation: Note that you are allowed to reuse a dictionary word.

Example 3:

Input:
s = "catsandog"
wordDict = ["cats", "dog", "sand", "and", "cat"]
Output:
[]

Solution:
　　方法一，使用递归：【通不过牛客的例子】

class Solution {
public:
    unordered_map<string, vector<string>>map;
    vector<string> wordBreak(string s, unordered_set<string> &dict) {
        if (map.find(s) != map.end())return map[s];
        if (s.empty())return { "" };
        vector<string>res;
        for (auto word : dict)
        {
            if (s.substr(0, word.size()) != word)continue;
            vector<string>rem = wordBreak(s.substr(word.size()), dict);
            for (auto str : rem)
                res.push_back(word + (str.empty() ? "" : " ") + str);
        }
        return map[s] = res;
    }
};

　　方法二：使用动态规划

//使用动态规划

class Solution {
public:
    vector<string> wordBreak(string s, unordered_set<string> &dict) {
        int len = s.length();
        dp = new vector<bool>[len];
        for (int pos = 0; pos < len; pos++) {
            for (int i = 1; i < len - pos + 1; i++) {
                if (dict.find(s.substr(pos, i)) != dict.end())
                    dp[pos].push_back(true);
                else
                    dp[pos].push_back(false);
            }
        }
        dfs(s, len - 1);
        return res;
    }
    void dfs(string s, int n) {
        if (n >= 0) {
            for (int i = n; i >= 0; i--) {
                if (dp[i][n - i]) {
                    mid.push_back(s.substr(i, n - i + 1));
                    dfs(s, i - 1);
                    mid.pop_back();
                }
            }
        }
        else {
            string r;
            for (int j = mid.size() - 1; j >= 0; j--) {
                r += mid[j];
                if (j > 0)
                    r += " ";
            }
            res.push_back(r);
        }
    }
    vector<bool> *dp;
    vector<string> res;
    vector<string> mid;
};