LeetCode 139. Word Break 原创Java参考解答
问题描述
https://leetcode.com/problems/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.
解题思路
题目是求一个字符串s是否能切分出所有wordDict里面的单词。
判断方案是否可行的题目则有很大可能是动态规划题，本题亦如此。
本题的动态规划方程f[i]： 表示s的前i个位置能否切分出wordDict里面的单词。初始化时，f[0]为true，表示空字符串””也是在wordDict里的单词。
二重循环，j < i，用来计算s字符串第 i 位置能否切出wordDict里面的单词。
- f[j] 用来获取s的前 j 个位置是否能切出wordDict里面的单词。
- 只有当f[j] = true（即s前 j 个字符能切出单词）并且s.substring(j, i) 也是在wordDict里面单词的时候，表明s的第 i 位置可以切分。
参考代码
public class Solution {
public boolean wordBreak(String s, List<String> wordDict) {
// f(x): whether the i th position of s is able to make word break
boolean[] f = new boolean[s.length() + 1];
f[0] = true;
for (int i = 0; i < s.length() + 1; i++) {
for (int j = 0; j < i; j++) {
if (!f[j]) {
continue;
}
String sub = s.substring(j, i);
if (wordDict.contains(sub)) {
f[i] = true;
break;
}
}
}
return f[s.length()];
}
}