[Swift]LeetCode859. 亲密字符串 | Buddy Strings
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?微信公眾號(hào):山青詠芝(shanqingyongzhi)
?博客園地址:山青詠芝(https://www.cnblogs.com/strengthen/)
?GitHub地址:https://github.com/strengthen/LeetCode
?原文地址:?https://www.cnblogs.com/strengthen/p/10588891.html?
?如果鏈接不是山青詠芝的博客園地址,則可能是爬取作者的文章。
?原文已修改更新!強(qiáng)烈建議點(diǎn)擊原文地址閱讀!支持作者!支持原創(chuàng)!
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Given two strings?A?and?B?of lowercase letters, return?true?if and only if we?can swap two letters in?A?so that the result equals?B.
Example 1:
Input: A = "ab", B = "ba" Output: trueExample 2:
Input: A = "ab", B = "ab" Output: falseExample 3:
Input: A = "aa", B = "aa" Output: trueExample 4:
Input: A = "aaaaaaabc", B = "aaaaaaacb" Output: trueExample 5:
Input: A = "", B = "aa" Output: falseNote:
給定兩個(gè)由小寫字母構(gòu)成的字符串?A?和?B?,只要我們可以通過交換?A?中的兩個(gè)字母得到與?B?相等的結(jié)果,就返回?true?;否則返回?false?。
示例 1:
輸入: A = "ab", B = "ba" 輸出: true示例 2:
輸入: A = "ab", B = "ab" 輸出: false示例 3:
輸入: A = "aa", B = "aa" 輸出: true示例 4:
輸入: A = "aaaaaaabc", B = "aaaaaaacb" 輸出: true示例 5:
輸入: A = "", B = "aa" 輸出: false提示:
16ms
?
1 class Solution { 2 func buddyStrings(_ A: String, _ B: String) -> Bool { 3 guard !A.isEmpty,!B.isEmpty, A.count == B.count else { 4 return false 5 } 6 if A == B && Set(A).count < A.count { return true } 7 let arrayA = Array(A) 8 let arrayB = Array(B) 9 var diffIndexArray:Array<Int> = [] 10 for i in 0..<A.count { 11 if arrayA[i] != arrayB[i] { 12 diffIndexArray.append(i) 13 } 14 if diffIndexArray.count > 2 { 15 return false 16 } 17 } 18 if diffIndexArray.count != 2 { return false } 19 return arrayA[diffIndexArray[1]] == arrayB[diffIndexArray[0]] && arrayA[diffIndexArray[0]] == arrayB[diffIndexArray[1]] 20 } 21 }20ms
1 class Solution { 2 func buddyStrings(_ As: String, _ Bs: String) -> Bool { 3 guard As.count == Bs.count && As.count > 1 else { 4 return false 5 } 6 if As == Bs { 7 if Set(As).count < As.count { 8 return true 9 } 10 return false 11 } 12 let A = Array(As) 13 let B = Array(Bs) 14 var diff = [Int]() 15 for i in 0..<A.count { 16 if A[i] != B[i] { 17 diff.append(i) 18 } 19 if diff.count > 2 { 20 return false 21 } 22 } 23 guard diff.count == 2 else { 24 return false 25 } 26 if A[diff[0]] == B[diff[1]] && A[diff[1]] == B[diff[0]] { 27 return true 28 } 29 return false 30 } 31 }24ms
1 class Solution { 2 func buddyStrings(_ A: String, _ B: String) -> Bool { 3 if A.length != B.length { return false } 4 if A.count < 2 || A.count != B.count { 5 return false 6 } 7 if A == B && Set(A).count < A.count { 8 return true 9 } 10 11 var a = Array(A) 12 var b = Array(B) 13 var array = [String]() 14 15 if A != B { 16 for i in 0 ..< a.count { 17 if a[i] != b[i] { 18 array.append(String(a[i])) 19 array.append(String(b[i])) 20 } 21 } 22 if array.reversed() == array { 23 return true 24 } 25 } 26 return false 27 } 28 }28ms
1 class Solution { 2 func buddyStrings(_ A: String, _ B: String) -> Bool { 3 if A.count != B.count { 4 return false 5 } 6 var map = [Character: Int]() 7 var x = -1 8 var y = -1 9 let arrayA = Array(A) 10 let arrayB = Array(B) 11 for i in 0..<A.count { 12 map[arrayA[i], default: 0] += 1 13 if arrayA[i] == arrayB[i] { 14 continue 15 } 16 if x == -1 { 17 x = i 18 } 19 else if y == -1 { 20 y = i 21 } 22 else { 23 return false 24 } 25 } 26 if x != -1 && y != -1 { 27 if arrayA[x] == arrayB[y] && arrayA[y] == arrayB[x] { 28 return true 29 } 30 } 31 else { 32 for (_, value) in map { 33 if value > 1 { 34 return true 35 } 36 } 37 } 38 return false 39 } 40 }32ms
1 class Solution { 2 func buddyStrings(_ A: String, _ B: String) -> Bool { 3 if A.length != B.length { return false } 4 if A.count < 2 || A.count != B.count { 5 return false 6 } 7 if A == B && Set(A).count < A.count { 8 return true 9 } 10 11 var a = Array(A).map({ String($0) }) 12 var b = Array(B).map({ String($0) }) 13 var arr = [String]() 14 15 if A != B { 16 for i in 0 ..< a.count { 17 if a[i] != b[i] { 18 arr.append(a[i]) 19 arr.append(b[i]) 20 } 21 } 22 if arr.reversed() == arr { return true } 23 } 24 25 return false 26 } 27 }40ms
1 class Solution { 2 func buddyStrings(_ A: String, _ B: String) -> Bool { 3 guard A.count == B.count else { 4 return false 5 } 6 var diff = [(Character, Character)]() 7 8 for (a, b) in zip(A, B) { 9 if a != b { 10 diff.append((a, b)) 11 if diff.count > 2 { 12 return false 13 } 14 } 15 } 16 17 return (diff.count == 0 && Set(Array(A)).count < A.count) || 18 (diff.count == 2 && diff[0] == (diff[1].1, diff[1].0)) 19 } 20 }?
轉(zhuǎn)載于:https://www.cnblogs.com/strengthen/p/10588891.html
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