标签：leetcode   算法   

问题描述：

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".

基本思想：

动态规划思想： 保存每个子串S（i，j）是否可分的信息。从小到大构建可分性表格。

代码：

    public boolean wordBreak(String s, Set<String> dict) {  //java
    	if(s.isEmpty())
    	    return true;
    	if(dict.contains(s))
    	    return true;
    	
    	int len = s.length();
    	int [][] record = new int[len+1][len+1];
    	for(int i=0; i<=len; i++)
    	    record[i][i]=1;
    	
    	for(int step=1; step<=len; step++)
    	{
        	for(int j=step; j<=len; j++)
        	{
        	    int i=j-step;
        	    if(dict.contains(s.substring(i,j)))
        	     {
        	         record[i][j]=1;
        	         continue;
        	     }
        	     
        	     for(int k=i+1; k<j; k++)
        	     {
        	         if(record[i][k]==1 && record[k][j]==1)
        	         {
        	            record[i][j] = 1;   
        	            break;
        	         }
        	     }
        	}
    	}
    
    	if(record[0][len] == 1) return true;
    	else return false;
    }

[leetcode]Word Break

标签：leetcode   算法   

原文地址：http://blog.csdn.net/chenlei0630/article/details/42468687