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(每日一练java)单词拆分 II

四月天2021 2022-02-19 阅读 75

单词拆分 II

给定一个非空字符串 s 和一个包含非空单词列表的字典 wordDict,在字符串中增加空格来构建一个句子,使得句子中所有的单词都在词典中。返回所有这些可能的句子。

说明:

  • 分隔时可以重复使用字典中的单词。
  • 你可以假设字典中没有重复的单词。

示例 1:

输入:
s = "catsanddog"
wordDict = ["cat", "cats", "and", "sand", "dog"]
输出:
[
  "cats and dog",
  "cat sand dog"
]

示例 2:

输入:
s = "pineapplepenapple"
wordDict = ["apple", "pen", "applepen", "pine", "pineapple"]
输出:
[
  "pine apple pen apple",
  "pineapple pen apple",
  "pine applepen apple"
]
解释: 注意你可以重复使用字典中的单词。

示例 3:

输入:
s = "catsandog"
wordDict = ["cats", "dog", "sand", "and", "cat"]
输出:
[]
class Solution {
    public List<String> wordBreak(String s, List<String> wordDict) {
        List<String> res = new ArrayList<>();
        int max = 0, min = Integer.MAX_VALUE;
        Set<String> set = new HashSet<>();
        for (String word : wordDict) {
            set.add(word);
            max = Integer.max(max, word.length());
            min = Integer.min(min, word.length());
        }
        boolean f[] = new boolean[s.length() + 1];
        f[0] = true;
        for (int i = 1; i < s.length() + 1; i++) {
            for (int j = Math.max(i - max, 0); j <= i - min; j++) {
                if (f[j] && set.contains(s.substring(j, i))) {
                    f[i] = true;
                    break;
                }
            }
        }
        if (f[s.length()]) {
            dfs(s, res, new StringBuilder(), set, 0, max, min);
        }
        return res;
    }
    private void dfs(String s, List<String> res, StringBuilder sb, Set<String> set, int index, int max, int min) {
        if (index == s.length()) {
            sb.deleteCharAt(sb.length() - 1);
            res.add(sb.toString());
            return;
        }
        String str;
        int size;
        for (int i = index + min; i <= s.length() && i <= index + max; i++) {
            if (set.contains(str = s.substring(index, i))) {
                size = sb.length();
                sb.append(str).append(' ');
                dfs(s, res, sb, set, i, max, min);
                sb.delete(size, sb.length());
            }
        }
    }
}

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