leetcode 522. Longest Uncommon Subsequence II

一。题目

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 beany 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].

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2.思路

每次拿出一个字符串和其他两两进行比较，判断是否有公共序列，若都没有公共序列则满足题目的要求，将该字符串长度与当前最大非公共子串比较，保留较大者。

3.代码

class Solution {public:    bool hasCommon(string a,string b){        int len1 = a.size();        int len2 = b.size();        while(len1>0&&len2>0){            int i = a.size()-len1;            int j = b.size()-len2;            if(a.at(i) == b.at(j)){                len1--;len2--;            }else{                len2--;            }        }        return len1==0;    }    int findLUSlength(vector<string>& strs) {        int maxLen = -1;        for(int i = 0;i<strs.size();++i){            int currentLen = strs[i].length();            bool all = true;            for(int j = 0;j<strs.size();++j){                if(i!=j&&hasCommon(strs[i], strs[j])){                    all = false;                    break;                }            }            if(all){                maxLen = maxLen<currentLen?currentLen:maxLen;            }        }        return maxLen;    }};