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

找最长不相同的子串。

1、满足条件的字符串，必须来自strs数组

2、该字符串不是数组strs数组中其他字符串的子串

3、找出满足上述条件，且长度最长的字符串

程序如下所示：

class Solution {
    public int findLUSlength(String[] strs) {
        int res = -1;
        for (int i = 0; i < strs.length; ++ i){
            int j = 0;
            for (j = 0; j < strs.length; ++ j){
                if (i == j){
                    continue;
                }
                if (strs[i].length() > strs[j].length()){
                    continue;
                }
                if (isSubsequence(strs[j], strs[i])){
                    break;
                }
            }
            if (j == strs.length){
                res = Math.max(res, strs[i].length());
            }
        }
        return res;
    }
    public boolean isSubsequence(String s0, String s1){
        int len0 = s0.length(), len1 = s1.length();
        int i = 0, j = 0;
        while (i < len0 && j < len1){
            if (s0.charAt(i) == s1.charAt(j)){
                i ++;
                j ++;
            }
            else {
                i ++;
            }
        }
        return j == len1;
    }
}