題目
Given a group of two strings, you need to find the longest uncommon subsequence of this group of two strings. 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 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.
Example 1:
Input: "aba", "cdc"
Output: 3
Explanation: The longest uncommon subsequence is "aba" (or "cdc"),
because "aba" is a subsequence of "aba",
but not a subsequence of any other strings in the group of two strings.
Note:
- Both strings' lengths will not exceed 100.
- Only letters from a ~ z will appear in input strings.
題目大意：
給定兩個(gè)字符串泉坐，計(jì)算其“最長不公共子序列”横殴。最長不公共子序列是指：兩字符串中某一個(gè)的子序列置媳，該子序列不是另一個(gè)字符串的子序列伊约，并且長度最長。
子序列是指從一個(gè)序列中刪除一些字符，剩余字符順序保持不變得到的新序列。任何字符串都是其本身的子序列盗扇，空串不屬于任意字符串的子序列。
返回最長不公共子序列沉填，若不存在疗隶，返回-1。
注意：
- 兩字符串長度均不超過100
- 輸入字符串只包含小寫字母a-z
解題思路：
若兩字符串不相等翼闹，選擇較長的字符串返回長度即可斑鼻。
否則返回-1。（若兩字符串相等猎荠，則任意字符串的子串均為另一個(gè)的子串）
代碼
findLUSlength.go
package _521_Longest_Uncommon_Subsequence_I
func FindLUSlength(a string, b string) int {
lena := len(a)
lenb := len(b)
if lena > lenb {
return lena
} else if lena < lenb {
return lenb
}else if a != b {
return lena
} else {
return -1
}
}
測試代碼
findLUSlength_test.go
package _521_Longest_Uncommon_Subsequence_I
import "testing"
func TestFindLUSlength(t *testing.T) {
input1 := "aba"
input2 := "cdc"
want := 3
ret := FindLUSlength(input1, input2)
if want == ret {
t.Logf("pass")
} else {
t.Errorf("fail, want %+v, get %+v", want, ret)
}
}
