140. Word Break II(hard)

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140. Word Break II

题目:

 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. You may assume the dictionary does not contain duplicate words.

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"].

UPDATE (2017/1/4):
The wordDict parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes. 

解析

  • unordered_set & dict办版本
    //运行时间:4ms
    //占用内存:508k

class Solution {

    vector<string> combine(string word, vector<string> prev){
        for (int i = 0; i < prev.size(); ++i){
            prev[i] += " " + word;
        }
        return prev;
    }

public:
    vector<string> wordBreak(string s, unordered_set<string>& dict) {
        
        vector<string> result;
        if (dict.count(s)){ //a whole string is a word
            result.push_back(s);
        }
        for (int i = 1; i < s.size(); ++i){
            string word = s.substr(i);
            if (dict.count(word)){
                string rem = s.substr(0, i);
                vector<string> prev = combine(word, wordBreak(rem, dict));
                result.insert(result.end(), prev.begin(), prev.end());
            }
        }
        
        reverse(result.begin(), result.end());
        return result;
    }
};
  • 暴力超时
namespace test
{
    vector<string> wordBreak(string s, unordered_set<string> &dict) {
        //暴力搜索,不能ac,复杂度超了。
        vector<string> res;
        int size = s.length();
        for (int i = 0; i < size; i++){
            string tmp = s.substr(0, i + 1);
            if (dict.count(tmp))
                findNext(res, s, dict, tmp, i + 1);
        }
        return res;
    }

    void findNext(vector<string> & res, string s, unordered_set<string> &dict, string tmp, int i){
        int size = s.length();
        if (i >= size){
            res.push_back(tmp);
            return;
        }
        for (int j = 1; j <= size - i; ++j){
            string now = s.substr(i, j);
            if (dict.count(now)){
                findNext(res, s, dict, tmp + ' ' + now, i + j);
            }
        }
    }
}

题目来源

    原文作者:ranjiewen
    原文地址: https://www.cnblogs.com/ranjiewen/p/8099147.html
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
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