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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".
解题思路一：
直接暴力枚举会导致TLE，因此，需要记录之前的结果，即可以采用dp的思路，JAVA实现如下：
static public boolean wordBreak(String s, Set<String> wordDict) {
boolean[] dp = new boolean[s.length() + 1];
dp[0] = true;
for (int i = 1; i < dp.length; i++)
for (int j = i; j >= 0 && !dp[i]; j--)
if (wordDict.contains(s.substring(i - j, i)))
dp[i] = dp[i - j];
return dp[dp.length - 1];
}
解题思路二：
考虑到下题用dp做不出来，暴力枚举肯定TLE，所以可以设置一个unmatch集合来存储s中已经确定无法匹配的子串，从而避免重复检查，JAVA实现如下：
static public boolean wordBreak(String s, Set<String> dict) {
return wordBreak(s, dict, new HashSet<String>());
}
static public boolean wordBreak(String s, Set<String> dict,
Set<String> unmatch) {
for (String prefix : dict) {
if (s.equals(prefix))
return true;
else if (s.startsWith(prefix)) {
String suffix = s.substring(prefix.length());
if (!unmatch.contains(suffix)) {
if (wordBreak(suffix, dict, unmatch))
return true;
else
unmatch.add(suffix);
}
}
}
return false;
}
Java for LeetCode 139 Word Break
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原文地址：http://www.cnblogs.com/tonyluis/p/4551146.html
