522. Longest Uncommon Subsequence II 

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class Solution {
    public int findLUSlength(String[] strs) {
        // 注意转换题意，如果一个序列为某字符串独有的子序列，那么最长的特殊序列一定为该序列所属的字符串

        int maxLength = -1;

        for (int i = 0; i < strs.length; i++) {
            if (strs[i].length() < maxLength) {
                continue;
            }

            boolean special = true;
            for (int j = 0; j < strs.length; j++) {
                if (i == j) {
                    continue;
                }
                if (isSubSequence(strs[i], strs[j])) {
                    // 如果是其他字符串的子序列，则不是特殊序列
                    special = false;
                    break;
                }
            }

            if (special) {
                maxLength = Math.max(maxLength, strs[i].length());
            }
        }

        return maxLength;
    }

    private boolean isSubSequence(String strA, String strB) {
        int m = strA.length(), n = strB.length();

        if (strB.length() < strA.length()) {
            return false;
        }    
        
        int[][] dp = new int[m + 1][n + 1];
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (strA.charAt(i - 1) == strB.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                } else {
                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }

        return dp[m][n] == strA.length();
    }
}

References

522. Longest Uncommon Subsequence II