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.

Note:

  • The same word in the dictionary may be reused multiple times in the segmentation.
  • You may assume the dictionary does not contain duplicate words.

Example 1:

Input: s = "leetcode", wordDict = ["leet", "code"]
Output: true
Explanation: Return true because "leetcode" can be segmented as "leet code".

Example 2:

Input: s = "applepenapple", wordDict = ["apple", "pen"]
Output: true
Explanation: Return true because "applepenapple" can be segmented as "apple pen apple".
             Note that you are allowed to reuse a dictionary word.

Example 3:

Input: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]
Output: false

单词拆分。题意是给一个String s和一个包含一些单词的list，请你返回用这些list里面的单词是否能拼接成S。

思路是DP，DP[i]的含义是以字母i结尾的字符串是否能被list中的单词拼接。初始化dp[0] = true。接下来用另外一个指针j去扫描0 - i范围内的substring。如果dp[j] = true && substring(j, i)也在wordDict存在，则dp[i] = true。

跑一下第三个例子，

Example 1:

Input: s = "leetcode", wordDict = ["leet", "code"]
Output: true
Explanation: Return true because "leetcode" can be segmented as "leet code".

DP数组最后的输出值如下，当i指针遍历到c（index = 4），j指针还在0的时候，此时因为dp[0] = true && s.substring(j, i) = s.substring(0, 4) = "leet"也存在于wordDict，所以可以将dp[4]标记为true。

[true, false, false, false, true, false, false, false, true]

时间O(n^2)

空间O(n)

Java实现

 1 class Solution {
 2     public boolean wordBreak(String s, List<String> wordDict) {
 3         boolean[] dp = new boolean[s.length() + 1];
 4         dp[0] = true;
 5         for (int i = 1; i <= s.length(); i++) {
 6             for (int j = 0; j < i; j++) {
 7                 if (dp[j] && wordDict.contains(s.substring(j, i))) {
 8                     dp[i] = true;
 9                     break;
10                 }
11             }
12         }
13         return dp[s.length()];
14     }
15 }

 

JavaScript实现

 1 /**
 2  * @param {string} s
 3  * @param {string[]} wordDict
 4  * @return {boolean}
 5  */
 6 var wordBreak = function (s, wordDict) {
 7     const dp = new Array(s.length + 1).fill(false);
 8     dp[0] = true;
 9     for (let i = 1; i <= s.length; i++) {
10         for (let j = 0; j < i; j++) {
11             const word = s.slice(j, i);
12             if (dp[j] == true && wordDict.includes(word)) {
13                 dp[i] = true;
14                 break;
15             }
16         }
17     }
18     return dp[s.length];
19 };

 

LeetCode 题目总结