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| 1 | +packagecom.stevesun.solutions; |
| 2 | + |
| 3 | +/** |
| 4 | + * Given a group of two strings, you need to find the longest uncommon subsequence of this group of two strings. |
| 5 | + * 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. |
| 6 | + * A subsequence is a sequence that can be derived from one sequence by deleting some characters without changing the order of the remaining elements. |
| 7 | + * Trivially, any string is a subsequence of itself and an empty string is a subsequence of any string. |
| 8 | + * The input will be two strings, and the output needs to be the length of the longest uncommon subsequence. If the longest uncommon subsequence doesn't exist, return -1. |
| 9 | +
|
| 10 | + Example 1: |
| 11 | + Input: "aba", "cdc" |
| 12 | + Output: 3 |
| 13 | + Explanation: The longest uncommon subsequence is "aba" (or "cdc"), |
| 14 | + because "aba" is a subsequence of "aba", |
| 15 | + but not a subsequence of any other strings in the group of two strings. |
| 16 | + Note: |
| 17 | +
|
| 18 | + Both strings' lengths will not exceed 100. |
| 19 | + Only letters from a ~ z will appear in input strings. |
| 20 | + */ |
| 21 | +publicclassLongestUncommonSubsequenceI { |
| 22 | +//The gotcha point of this question is: |
| 23 | +//1. if a and b are identical, then there will be no common subsequence, return -1 |
| 24 | +//2. else if a and b are of equal length, then any one of them will be a subsequence of the other string |
| 25 | +//3. else if a and b are of different length, then the longer one is a required subsequence because the longer string cannot be a subsequence of the shorter one |
| 26 | +publicintfindLUSlength(Stringa,Stringb) { |
| 27 | +if (a.equals(b))return -1; |
| 28 | +returnMath.max(a.length(),b.length()); |
| 29 | + } |
| 30 | + |
| 31 | +} |