Description

You are given an array of words where each word consists of lowercase English letters.

wordA is a predecessor of wordB if and only if we can insert exactly one letter anywhere in wordA without changing the order of the other characters to make it equal to wordB.

For example, “abc” is a predecessor of “abac”, while “cba” is not a predecessor of “bcad”.
A word chain is a sequence of words [word1, word2, …, wordk] with k >= 1, where word1 is a predecessor of word2, word2 is a predecessor of word3, and so on. A single word is trivially a word chain with k == 1.

Return the length of the longest possible word chain with words chosen from the given list of words.

Examples

Example 1:

Input: words = [“a”,“b”,“ba”,“bca”,“bda”,“bdca”]
Output: 4
Explanation: One of the longest word chains is [“a”,“ba”,“bda”,“bdca”].

Example 2:

Input: words = [“xbc”,“pcxbcf”,“xb”,“cxbc”,“pcxbc”]
Output: 5
Explanation: All the words can be put in a word chain [“xb”, “xbc”, “cxbc”, “pcxbc”, “pcxbcf”].

Example 3:

Input: words = [“abcd”,“dbqca”]
Output: 1
Explanation: The trivial word chain [“abcd”] is one of the longest word chains.
[“abcd”,“dbqca”] is not a valid word chain because the ordering of the letters is changed.

Constraints:

1 <= words.length <= 1000
1 <= words[i].length <= 16
words[i] only consists of lowercase English letters.

思路

思路就是先对words进行排序，我这里定义的排序规则是按照字符串长度进行排序，这样在后面进行dp数组更新的时候会比较和谐。

排序之后会得到一个字符串长度升序的序列，然后就 O ( n 2 ) O(n^2) O(n2)判断是否两个string为predecessor关系

代码

class Solution {
    public Boolean isPreDecessor(String a, String b){
        if (a.charAt(0) != b.charAt(0))
            return a.substring(1).equals(b);
        if (a.charAt(a.length() - 1) != b.charAt(b.length() - 1))
            return a.substring(0, a.length() - 1).equals(b);
        
        int count = 0;
        for(int i = 0; i < a.length() - 1; i++){
            if (a.charAt(i) != b.charAt(i))
                break;
            count++;
        }
        for(int i = 0; i < a.length() - 1; i++){
            if (a.charAt(a.length() - 1 - i) != b.charAt(b.length() - 1 - i))
                break;
            count++;
        }
        
        return count >= b.length();
    }
    
    public int longestStrChain(String[] words) {
        Arrays.sort(words, new Comparator<String>(){
            @Override
            public int compare(String o1, String o2) {
                return o1.length() - o2.length();
            }
        });
        
        int[] chainLength = new int[words.length];
        for(int i = 0; i < words.length; i++){
            chainLength[i] = 1;
            for (int j = 0; j < i; j++){
                if (words[i].length() - words[j].length() == 1 && isPreDecessor(words[i], words[j]))
                    chainLength[i] = Math.max(chainLength[j] + 1, chainLength[i]);
            }
        }
        
        int maxLen = 0;
        for (int i = 0; i < words.length; i++)
            maxLen = Math.max(maxLen, chainLength[i]);
        
        return maxLen;
    }
}