题目描述：

LeetCode 522. Longest Uncommon Subsequence II

Given a list of strings, you need to find the longest uncommon subsequence among them. The longest uncommon subsequence is defined as the longest subsequence of one of these strings and this subsequence should not be any subsequence of the other strings.

A subsequence is a sequence that can be derived from one sequence by deleting some characters without changing the order of the remaining elements. Trivially, any string is a subsequence of itself and an empty string is a subsequence of any string.

The input will be a list of strings, and the output needs to be the length of the longest uncommon subsequence. If the longest uncommon subsequence doesn't exist, return -1.

Example 1:

Input: "aba", "cdc", "eae"
Output: 3

Note:

1. All the given strings' lengths will not exceed 10.
2. The length of the given list will be in the range of [2, 50].

题目大意：

给定一组字符串，寻找其最长不公共子序列。最长不公共子序列是指：这组字符串中某一个的子序列，该子序列不是其余任意字符串的子序列，并且长度最长。

子序列是指从一个序列中删除一些字符，剩余字符顺序保持不变得到的新序列。任何字符串都是其本身的子序列，空串不属于任意字符串的子序列。

返回最长不公共子序列，若不存在，返回-1。

注意：

1. 两字符串长度均不超过10
2. 给定字符串列表长度范围为[2, 50]

解题思路：

首先将输入字符串列表strs按照长度递减排序，记得到的新列表为slist。

利用计数器cnt统计每个字符串出现的次数。

遍历slist，记当前字符串为c，其下标为i：

    若c在strs中出现不止一次，跳过该字符串

    否则，利用贪心算法对c和slist[0 .. i - 1]的字符串进行匹配，若均匹配失败，则返回len(c)

遍历结束，返回-1

Python代码：

class Solution(object):
    def uncommon(self, parent, child):
        lp, lc = len(parent), len(child)
        pp = pc = 0
        while pp < lp and pc < lc:
            if parent[pp] == child[pc]:
                pc += 1
            pp += 1
        return pc != lc
    def findLUSlength(self, strs):
        """
        :type strs: List[str]
        :rtype: int
        """
        cnt = collections.Counter(strs)
        slist = sorted(set(strs), key=len, reverse=True)
        for i, c in enumerate(slist):
            if cnt[c] > 1: continue
            if all(self.uncommon(p, c) for p in slist[:i]):
                return len(c)
        return -1

 

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