LeetCode——Word Break

标签：leetcode

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".

原题链接：https://oj.leetcode.com/problems/word-break/

题目：给定一个字符串和一个单词词典，判断字符串能否被切分成空格隔开的存在于词典中的单词序列。

思路：从头开始依次一个字符地向后截取并判断。逻辑应该是对的。测试超时了。

	public static boolean wordBreak1(String s, Set<String> dict) {
		if(dict.size() <= 0)
			return false;
		int len = s.length();
		if(len <= 0)
			return true;
		boolean flag = false;
		for(int i=1;i<=len;i++){
			String tmp = s.substring(0, i);
			if(dict.contains(tmp)){
				if(tmp.length() == len)
					return true;
			}
			flag = wordBreak(s.substring(i),dict);
		}
		return flag;
	}

	//dp[i] 代表 字符串(j,i)能被分词否
	public static boolean wordBreak(String s, Set<String> dict) {
		int length = s.length();
		boolean[] dp = new boolean[length + 1];
		dp[0] = true;
		for (int i = 1; i <= length; i++) {
			for (int j = 0; j < i; j++) {
				if (dp[j] && dict.contains(s.substring(j, i))) {
					dp[i] = true;
					break;
				}
			}
		}
		return dp[length];
	}

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LeetCode——Word Break

标签：leetcode

原文地址：http://blog.csdn.net/laozhaokun/article/details/37600733