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].

這道題讓找最長的獨有子序列，解題思路可以分三步：

1、按照字元串長度降序排列strs

2、周遊strs，如果str不是所有strs的獨有子字元串，傳回str的長度

3、如果沒有找到獨有字元串，傳回-1

按照解題思路寫代碼，代碼如下：

public class Solution {
    public int findLUSlength(String[] strs) {
        Arrays.sort(strs, new Comparator<String>() {
            public int compare(String s1, String s2) {
                return s2.length() - s1.length();
            }
        });
        for (int i = 0; i < strs.length; i ++) {
            int noMatches = strs.length - 1;
            for (int j = 0; j < strs.length; j ++) {
                if (i != j && isSubString(strs[i], strs[j])) {
                    noMatches --;
                }
                if (noMatches == 0) {
                    return strs[i].length();
                }
            }
        }
        return -1;
    }
    
    private boolean isSubString(String s1, String s2) {
        int i = 0;
        for (char ch: s2.toCharArray()) {
            if (i < s1.length() && s1.charAt(i) == ch) {
                i ++;
            }
        }
        if (i == s1.length()) {
            return false;
        }
        return true;
    }
}