140. Word Break II

最新推荐文章于 2021-07-24 20:33:45 发布

无差别刷题

最新推荐文章于 2021-07-24 20:33:45 发布

阅读量94

点赞数

CC 4.0 BY-SA版权

文章标签： recursion dynamic programming backtracking leetcode

本文链接：https://blog.youkuaiyun.com/wilzxu/article/details/89172184

本文探讨了WordBreak II问题的解决策略，采用递归与哈希表优化的方法，实现字符串分割成字典中单词的所有可能句子组合。通过实例演示算法流程，分析时间与空间复杂度。

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:
[]

方法1:

huahua: https://www.youtube.com/watch?v=JqOIRBC0_9c
思路：

和139. Word Break 一样的思路，不一样的是这里要用hash把递归解全部记下来，与上一层合并。在helper里，首先要剪枝所有查询并记录过的结果，如果找到了立刻返回。否则开始递归。递归过程中首先建立一个新的ans来暂时保存所有当前s的解集。1. 要查整个s是否在dict里，是的话先推入，2. 左边分割出1个，2个，… size() - 1个字符，一次查询右边是否inDict，如果没有可以立刻结束本循环尝试下一种分割。而如果有，需要开始递归解左边。如果假设左边的所有解被返回装进了left_ans，那么当前s的解集包括所有string s: left_ans + right的组合，i.e. 个数由左边解集大小决定，同时这样也包括了左边不可解的情况，因为自动返回空集。append是一个util函数，用来完成这一步追加。注意，此时不能终止循环，因为查询所有解，必须将所有分割点尝试一遍才能最终送入hash，否则解不完整且重复多次被送入。

Complexity

Time complexity: O(2^n)
Space complexity: O(2^n)

class Solution {
public:
    vector<string> wordBreak(string s, vector<string>& wordDict) {
        unordered_set<string> wordSet(wordDict.begin(), wordDict.end());
        return breakHelper(s, wordSet);
    }
private:
    unordered_map<string, vector<string> > hash;
    void append(vector<string> & left_ans, string & right){
        for (string & s: left_ans){
            s += " " + right;
        }
    }    
    
    vector<string> breakHelper(string & s, unordered_set<string> & wordSet){
        if (hash.count(s)) return hash[s];
        vector<string> ans;
        if (wordSet.count(s)) ans.push_back(s);
        for (int i = 1; i < s.size(); i++){
            string left = s.substr(0, i);
            string right = s.substr(i);
            if (wordSet.count(right)) {
                vector<string> left_ans = breakHelper(left, wordSet);
                append(left_ans, right);
                ans.insert(ans.end(), left_ans.begin(), left_ans.end());
            }
        }
        hash[s] = ans;
        return hash[s];
    }
};