140. Word Break II（js）

最新推荐文章于 2021-05-30 21:00:13 发布

转载最新推荐文章于 2021-05-30 21:00:13 发布 · 91 阅读

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原文链接：http://www.cnblogs.com/xingguozhiming/p/11006721.html

本文解析了LeetCode上编号为140的题目“Word Break II”，详细介绍了如何通过动态规划算法解决字符串切割问题，使得切割后的每个子串都是给定字典中的有效单词。

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140. Word Break II

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences.

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 = "catsanddog"
wordDict = ["cat", "cats", "and", "sand", "dog"]
Output:
[
  "cats and dog",
  "cat sand dog"
]

Example 2:

Input:
s = "pineapplepenapple"
wordDict = ["apple", "pen", "applepen", "pine", "pineapple"]
Output:
[
  "pine apple pen apple",
  "pineapple pen apple",
  "pine applepen apple"
]
Explanation: Note that you are allowed to reuse a dictionary word.

Example 3:

Input:
s = "catsandog"
wordDict = ["cats", "dog", "sand", "and", "cat"]
Output:
[]
题意：将字符串切割，求出所有情况满足子串都在wordDict中存在

/**
 * @param {string} s
 * @param {string[]} wordDict
 * @return {string[]}
 */
var wordBreak = function(s, wordDict) {
    var res = [];
    var from = [];
    from[0] = [0];
    var wordDict=new Set(wordDict);
    for (var i = 1; i <= s.length; i++) {
        from[i] = [];
        for (var j = 0; j < i; j++) {
          if (from[j].length && wordDict.has(s.substring(j, i))) {
            from[i].push(j);
          }
        }
    }
    build(s.length, '');
    return res;

    function build(idx, suffix) {
        if (!idx) return res.push(suffix);
        from[idx].forEach(function(from) {
          build(from, suffix === '' ? s.substring(from, idx) : s.substring(from, idx) + ' ' + suffix);
        })
    }
}

出处：https://leetcode.com/problems/word-break-ii/discuss/44384/Share-my-DP-JavaScript-solution

转载于:https://www.cnblogs.com/xingguozhiming/p/11006721.html