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1 | 1 | packagecom.fishercoder.solutions; |
2 | 2 |
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3 | 3 | /** |
| 4 | + * 521. Longest Uncommon Subsequence I |
| 5 | + * |
4 | 6 | * Given a group of two strings, you need to find the longest uncommon subsequence of this group of two strings. |
5 | 7 | * 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 | 8 | * 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 | 9 | * 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. |
| 10 | + * The input will be two strings, and the output needs to be the length of the longest uncommon subsequence. |
| 11 | + * If the longest uncommon subsequence doesn't exist, return -1. |
9 | 12 |
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10 | 13 | Example 1: |
11 | 14 | Input: "aba", "cdc" |
12 | 15 | Output: 3 |
13 | 16 | Explanation: The longest uncommon subsequence is "aba" (or "cdc"), |
14 | 17 | because "aba" is a subsequence of "aba", |
15 | 18 | but not a subsequence of any other strings in the group of two strings. |
16 | | - Note: |
17 | 19 |
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| 20 | + Note: |
18 | 21 | Both strings' lengths will not exceed 100. |
19 | 22 | Only letters from a ~ z will appear in input strings. |
20 | 23 | */ |
21 | 24 | publicclass_521 { |
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 | | - |
27 | | -//Or in other words, when a.length() != b.length(), no subsequence of b will be equal to a, so return Math.max(a.length(), b.length()) |
28 | | -publicintfindLUSlength(Stringa,Stringb) { |
29 | | -if (a.equals(b)) { |
30 | | -return -1; |
| 25 | +publicstaticclassSolution1 { |
| 26 | +/** |
| 27 | + * The gotcha point of this question is: |
| 28 | + * 1. if a and b are identical, then there will be no common subsequence, return -1 |
| 29 | + * 2. else if a and b are of equal length, then any one of them will be a subsequence of the other string |
| 30 | + * 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 |
| 31 | + * Or in other words, when a.length() != b.length(), no subsequence of b will be equal to a, so return Math.max(a.length(), b.length()) |
| 32 | + */ |
| 33 | +publicintfindLUSlength(Stringa,Stringb) { |
| 34 | +if (a.equals(b)) { |
| 35 | +return -1; |
| 36 | + } |
| 37 | +returnMath.max(a.length(),b.length()); |
31 | 38 | } |
32 | | -returnMath.max(a.length(),b.length()); |
33 | 39 | } |
34 | | - |
35 | 40 | } |