[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.

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

class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        boolean[] dp = new boolean[s.length() + 1];
        dp[0] = true;
        for (int i = 1; i <= s.length(); i++) {
            for (int j = 0; j < i; j++) {
                if (dp[j] && wordDict.contains(s.substring(j, i))) {
                    dp[i] = true;
                    break;
                }
            }
        }
        return dp[s.length()];
    }
}

JavaScript实现

/**
 * @param {string} s
 * @param {string[]} wordDict
 * @return {boolean}
 */
var wordBreak = function (s, wordDict) {
    const dp = new Array(s.length + 1).fill(false);
    dp[0] = true;
    for (let i = 1; i <= s.length; i++) {
        for (let j = 0; j < i; j++) {
            const word = s.slice(j, i);
            if (dp[j] == true && wordDict.includes(word)) {
                dp[i] = true;
                break;
            }
        }
    }
    return dp[s.length];
};

LeetCode 题目总结