LeetCode刷题笔录Word Break II

最新推荐文章于 2019-04-20 21:21:48 发布

原创最新推荐文章于 2019-04-20 21:21:48 发布 · 703 阅读

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本文介绍了一种使用回溯算法解决字符串拆分问题的方法，通过递归地检查子串是否存在于字典中来构造所有可能的句子。

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"].

这题在道理上和word break没啥区别，不一样的是这里要存储所有的中间结果。如果还用动态规划的话有一点麻烦。可以像N-Queens一样用暴力方法一个一个搜索。

public class Solution {
    public List<String> wordBreak(String s, Set<String> dict) {
        List<String> res = new ArrayList<String>();
        if(s.length() == 0 || s == null)
            return res;
        recursive(s, 0, dict, "", res);
        return res;
    }
    
    public void recursive(String s, int index, Set<String> dict, String string, List<String> res){
        if(index >= s.length()){
            res.add(string);
            return;
        }
        StringBuilder builder = new StringBuilder();
        for(int i = index; i < s.length(); i++){
            builder.append(s.charAt(i));
            if(dict.contains(builder.toString())){
                String str = string.isEmpty() ? builder.toString() : (string + " " + builder.toString());
                recursive(s, index + 1, dict, str, res);
            }
        }
    }
}

不过这里有一个比较变态的test case过不去