0820. Short Encoding of Words (M)

题目

A valid encoding of an array of words is any reference string s and array of indices indices such that:

• words.length == indices.length
• The reference string s ends with the '#' character.
• For each index indices[i], the substring of s starting from indices[i] and up to (but not including) the next '#' character is equal to words[i].

Given an array of words, return the length of the shortest reference string s possible of any valid encoding of words.

Example 1:

Input: words = ["time", "me", "bell"]
Output: 10
Explanation: A valid encoding would be s = "time#bell#" and indices = [0, 2, 5].
words[0] = "time", the substring of s starting from indices[0] = 0 to the next '#' is underlined in "time#bell#"
words[1] = "me", the substring of s starting from indices[1] = 2 to the next '#' is underlined in "time#bell#"
words[2] = "bell", the substring of s starting from indices[2] = 5 to the next '#' is underlined in "time#bell#"

Example 2:

Input: words = ["t"]
Output: 2
Explanation: A valid encoding would be s = "t#" and indices = [0].

Constraints:

• 1 <= words.length <= 2000
• 1 <= words[i].length <= 7
• words[i] consists of only lowercase letters.

题意

将数组中的元素按照指定规则拼成一个最短的字符串。

思路

主要是判断一个字符串是否是另一个字符串的后缀。可以这样处理，先将数组中所有字符串逆置，再按照字典序从大到小排序，这样前缀重合的字符串一定排在一起，且较长的排在前面；遍历数组，如果一个字符串是另一个字符串的前缀，则删去短串，根据最后留下的集合计算答案。

代码实现

Java

class Solution {
    public int minimumLengthEncoding(String[] words) {
        List<String> transform = new ArrayList<>();
        for (String word : words) {
            transform.add(new StringBuilder(word).reverse().toString());
        }
        Collections.sort(transform, Comparator.reverseOrder());
        int i = 0, j = 1;
        int count = 0, len = 0;
        while (j < transform.size()) {
            if (!transform.get(i).startsWith(transform.get(j))) {
                count++;
                len += transform.get(i).length();
                i = j;
            }
            j++;
        }
        count++;
        len += transform.get(i).length();
        return len + count;
    }
}