leetcode:Word Break II

最新推荐文章于 2020-06-04 11:37:52 发布

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最新推荐文章于 2020-06-04 11:37:52 发布

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本文链接：https://blog.youkuaiyun.com/chanrenyuan/article/details/41214767

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本文探讨了如何解决给定字符串与字符串集合匹配并组合的问题，通过实例演示了解决过程和算法实现，提供了完整的代码示例，适用于字符串处理和组合算法的学习与应用。

给出一个字符串，和一个字符串集合，问该字符串是否可以有集合中的字符串组成

Word Break II

Total Accepted: 20538 Total Submissions: 123580 My Submissions

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"].

因为这个例子超时了

Last executed input:

"aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab", ["a","aa","aaa","aaaa","aaaaa","aaaaaa","aaaaaaa","aaaaaaaa","aaaaaaaaa","aaaaaaaaaa"]

显然b不是集合里面的值，所以预先判断了一下，是否有解

import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

public class Solution {

    static List<String> ans;
    static char[] ch;
    static Set<String> set;
    
    public List<String> wordBreak(String s, Set<String> dict) { 
        ans = new ArrayList<String>();
        ch = s.toCharArray();
        set = dict;
        boolean flag = false;
        for(int i = ch.length - 1; i >= 0; --i){
            String temp = new String(ch, i, ch.length - i);
            if(set.contains(temp)){
                flag = true;
                break;
            }
        }
        if(flag)
            for(int i = 0; i < ch.length; ++i){
                String tmp = new String(ch, 0, i + 1);
                if(set.contains(tmp))dfs(i + 1, tmp);
            }
        return ans;
    }
    
    public static void dfs(int pre, String s){
        if(pre == ch.length){
            ans.add(s);
        }else if(pre < ch.length){
            for(int i = pre; i < ch.length; ++i){
                String temp = new String(ch, pre, i - pre + 1);
                if(set.contains(temp)){
                    dfs(i + 1, s + " " + temp);
                }
            }
        }
    }

    public static void main(String[] args) {
        String string = "catsanddog";
        HashSet<String> set = new HashSet<>();
        for (String str : new String[]{"cat", "cats", "and", "sand", "dog"}) {
            set.add(str);
        }
        Solution s = new Solution();
        System.out.println(s.wordBreak(string, set));
    }
}