leetcode--Word Break

缺乏、安全感 2021-10-25 12:42 244阅读 0赞

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