leetcode--Word Break

转载于 2019-08-20 23:30:01 发布 · 81 阅读

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CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/averillzheng/p/3571147.html

本文介绍了一种使用动态规划解决字符串分割问题的方法，通过判断一个字符串是否能被分割为字典中的单词序列。文章提供了两种实现方式，详细展示了如何利用布尔数组记录子问题的解决方案，从而高效地解决整体问题。

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Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

For example, given
s = "leetcode",
dict = ["leet", "code"].

Return true because "leetcode" can be segmented as "leet code".

Have you been asked this question in an interview?

dynamic programming method

public class Solution {
    public boolean wordBreak(String s, Set<String> dict) {
       if(s.length() <= 0 || dict.size() <= 0)
			return false;
		//check if s.substring(0, i) can be split
		boolean[] split = new boolean[s.length() + 1];
		split[0] = false;
		for(int i = 1; i < s.length() + 1; ++i){
			if(dict.contains(s.substring(0, i)))
				split[i] = true;
			else {
				for(int j = 0; j < i; ++j) {
					if(! split[j]) continue;
					split[i] = split[j] && dict.contains(s.substring(j, i));
					if(split[i])
						break;
				}	
			}
		}		
		return split[s.length()];
    }
}

second code:

public class Solution {
    public boolean wordBreak(String s, Set<String> dict) {
       if(s == null || dict.size() <= 0)
			return false;
	
		int len = s.length();
		
		boolean [][] splits = new boolean[len][len + 1];
		splits[0][0] = false;
		for(int i = len - 1; i >= 0; --i){
			splits[i][i] = false;
			for(int j = i + 1; j <= len; ++j){
				if(dict.contains(s.substring(i,j)))
					splits[i][j] = true;
				else{
					for(int k = i; k < j; ++k){
						splits[i][j] = splits[i][k] && splits[k][j];
						if(splits[i][j])
							break;
					}				
				}
			}
		}		
		return splits[0][len];
    }
}

转载于:https://www.cnblogs.com/averillzheng/p/3571147.html