144 lines
4.0 KiB
Markdown
144 lines
4.0 KiB
Markdown
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## 题目地址
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https://leetcode.com/problems/word-break/description/
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## 题目描述
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```
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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.
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Note:
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The same word in the dictionary may be reused multiple times in the segmentation.
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You may assume the dictionary does not contain duplicate words.
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Example 1:
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Input: s = "leetcode", wordDict = ["leet", "code"]
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Output: true
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Explanation: Return true because "leetcode" can be segmented as "leet code".
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Example 2:
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Input: s = "applepenapple", wordDict = ["apple", "pen"]
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Output: true
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Explanation: Return true because "applepenapple" can be segmented as "apple pen apple".
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Note that you are allowed to reuse a dictionary word.
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Example 3:
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Input: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]
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Output: false
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```
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## 思路
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这道题是给定一个字典和一个句子，判断该句子是否可以由字典里面的单词组出来，一个单词可以用多次。
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暴力的方法是无解的，复杂度极其高。 我们考虑其是否可以拆分为小问题来解决。
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对于问题`(s, wordDict)` 我们是否可以用(s', wordDict) 来解决。 其中s' 是s 的子序列，
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当s'变成寻常(长度为0)的时候问题就解决了。 我们状态转移方程变成了这道题的难点。
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我们可以建立一个数组dp, dp[i]代表 字符串 s.substring(0, i) 能否由字典里面的单词组成，
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值得注意的是，这里我们无法建立dp[i] 和 dp[i - 1] 的关系，
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我们可以建立的是dp[i - word.length] 和 dp[i] 的关系。
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我们用图来感受一下：
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没有明白也没有关系，我们分步骤解读一下：
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（以下的图左边都代表s，右边都是dict，灰色代表没有处理的字符，绿色代表匹配成功，红色代表匹配失败）
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上面分步解释了算法的基本过程，下面我们感性认识下这道题，我把它比喻为
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你正在`往一个老式手电筒🔦中装电池`
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## 代码
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```js
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/*
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* @lc app=leetcode id=139 lang=javascript
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*
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* [139] Word Break
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*
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* https://leetcode.com/problems/word-break/description/
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*
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* algorithms
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* Medium (34.45%)
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* Total Accepted: 317.8K
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* Total Submissions: 913.9K
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* Testcase Example: '"leetcode"\n["leet","code"]'
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*
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* Given a non-empty string s and a dictionary wordDict containing a list of
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* non-empty words, determine if s can be segmented into a space-separated
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* sequence of one or more dictionary words.
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*
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* Note:
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*
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*
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* The same word in the dictionary may be reused multiple times in the
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* segmentation.
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* You may assume the dictionary does not contain duplicate words.
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*
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*
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* Example 1:
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*
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*
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* Input: s = "leetcode", wordDict = ["leet", "code"]
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* Output: true
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* Explanation: Return true because "leetcode" can be segmented as "leet
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* code".
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*
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*
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* Example 2:
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*
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*
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* Input: s = "applepenapple", wordDict = ["apple", "pen"]
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* Output: true
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* Explanation: Return true because "applepenapple" can be segmented as "apple
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* pen apple".
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* Note that you are allowed to reuse a dictionary word.
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*
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*
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* Example 3:
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*
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*
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* Input: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]
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* Output: false
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*
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*
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*/
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/**
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* @param {string} s
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* @param {string[]} wordDict
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* @return {boolean}
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*/
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var wordBreak = function(s, wordDict) {
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const dp = Array(s.length + 1);
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dp[0] = true;
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for (let i = 0; i < s.length + 1; i++) {
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for (let word of wordDict) {
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if (dp[i - word.length] && word.length <= i) {
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if (s.substring(i - word.length, i) === word) {
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dp[i] = true;
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}
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}
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}
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}
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return dp[s.length] || false;
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};
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```
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