leetcode 139. Word Break

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words. You may assume the dictionary does not contain duplicate words.

For example, given

s = “leetcode”,

dict = [“leet”, “code”].

Return true because “leetcode” can be segmented as “leet code”.

UPDATE (2017/1/4):

The wordDict parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.

判断字符串能不能用字典中的字符串分割完。

思路：动态规划，从左至右开始匹配字符串，dp[i]表示字符串能够被分割至第i个位置，所以结果就是返回dp[s.length]

public boolean wordBreak(String s, List<String> wordDict) {
if(s == null || s.equals("")){
return true;
}
boolean[] dp = new boolean[s.length()+1];
dp[0] = true;
for(int i =0;i<s.length();i++){
for(String str : wordDict){
if(dp[i] && s.substring(i).startsWith(str)){
dp[i+str.length()] = true;
if(i+str.length()==s.length()){
return true;
}
}

}

}
return false;
}

运行结果

9ms，超过了92.46%的提交