You are given a 0-indexed string array words, where words[i] consists of lowercase English letters.
In one operation, select any index i such that 0 < i < words.length and words[i - 1] and words[i] are anagrams, and delete words[i] from words. Keep performing this operation as long as you can select an index that satisfies the conditions.
Return words after performing all operations. It can be shown that selecting the indices for each operation in any arbitrary order will lead to the same result.
An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase using all the original letters exactly once. For example, "dacb" is an anagram of "abdc".
Example 1:
Input: words = ["abba","baba","bbaa","cd","cd"]
Output: ["abba","cd"]
Explanation:
One of the ways we can obtain the resultant array is by using the following operations:
- Since words[2] = "bbaa" and words[1] = "baba" are anagrams, we choose index 2 and delete words[2].
Now words = ["abba","baba","cd","cd"]. - Since words[1] = "baba" and words[0] = "abba" are anagrams, we choose index 1 and delete words[1].
Now words = ["abba","cd","cd"]. - Since words[2] = "cd" and words[1] = "cd" are anagrams, we choose index 2 and delete words[2].
Now words = ["abba","cd"].
We can no longer perform any operations, so ["abba","cd"] is the final answer.
Example 2:
Input: words = ["a","b","c","d","e"]
Output: ["a","b","c","d","e"]
Explanation:
No two adjacent strings in words are anagrams of each other, so no operations are performed.
Constraints:
1 <= words.length <= 100
1 <= words[i].length <= 10
words[i] consists of lowercase English letters.
移除字母异位词后的结果数组。
给你一个下标从 0 开始的字符串 words ，其中 words[i] 由小写英文字符组成。在一步操作中，需要选出任一下标 i ，从 words 中 删除 words[i] 。其中下标 i 需要同时满足下述两个条件：
0 < i < words.length
words[i - 1] 和 words[i] 是 字母异位词 。
只要可以选出满足条件的下标，就一直执行这个操作。在执行所有操作后，返回 words 。可以证明，按任意顺序为每步操作选择下标都会得到相同的结果。
字母异位词 是由重新排列源单词的字母得到的一个新单词，所有源单词中的字母通常恰好只用一次。例如，"dacb" 是 "abdc" 的一个字母异位词。
思路
从左到右遍历 words，维护上一次「保留下来的单词」，记为 prev。当前单词 cur 与 prev 相同则为相邻变位词，跳过；否则加入答案并更新 prev。
复杂度
时间O(n)
空间O(k * klogk) - k是words的长度，klogk是排序的复杂度
代码
Java实现
class Solution {public List<String> removeAnagrams(String[] words) {List<String> res = new ArrayList<>();String prev = ""; // 上一个保留单词的排序签名for (String word : words) {String sorted = helper(word);if (!sorted.equals(prev)) { // 如果不是变位词res.add(word); // 加入原单词prev = sorted; // 更新签名}}return res;}private String helper(String word) {char[] letters = word.toCharArray();Arrays.sort(letters);return new String(letters);}
}