522. Longest Uncommon Subsequence II

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].

My Solutions:

1。初始化ans=-1

2。i从0到strs.length，检查strs[i]是不是别的string的subseqence

2。1。检查strs里除了i的每一个string，如果strs[i]是任意一个别的string的subsequence，就不用接着检查了，isSub=true, break出inner for loop

2。2。如果检查到结尾，发现strs[i]仍然不是任意一个别的string的subsequence，更新ans的长度

class Solution {
    public int findLUSlength(String[] strs) {
        int ans = -1;
        
        for (int i = 0; i < strs.length; i++) {
            boolean isSub = false;
            for (int j = 0; j < strs.length; j++) {
                if (i != j && isSubsequence(strs[i], strs[j])) {
                    isSub = true;
                    break;
                }
            }
            if (!isSub) ans = Math.max(ans, strs[i].length());
        }
        return ans;
    }
    
    // check whether a is b's subsequence
    public boolean isSubsequence(String a, String b) {
        
        if (a.length() > b.length()) return false;
        if (a.equals(b)) return true;
        
        int index = 0;
        for (int i = 0; i < b.length() && index < a.length(); i++) {
            if (a.charAt(index) == b.charAt(i)) index++;
        }
        return index == a.length();
    }
}