[Swift]LeetCode888.公平的糖果交换|FairCandySwap

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Alice and Bob have candy bars of different sizes: A[i] is the size of the i-th bar of candy that Alice has, and B[j] is the size of the j-th bar of candy that Bob has.

Since they are friends, they would like to exchange one candy bar each so that after the exchange, they both have the same total amount of candy. (The total amount of candy a person has is the sum of the sizes of candy bars they have.)

Return an integer array ans where ans[0] is the size of the candy bar that Alice must exchange, and ans[1] is the size of the candy bar that Bob must exchange.

If there are multiple answers, you may return any one of them. It is guaranteed an answer exists.

Example 1:

Input: A = [2,2]
Output: [1,2]

Example 2:

Input: A = [2,3]
Output: [1,2]

Example 3:

Input: A = [1,3]
Output: [2,3]

Example 4:

Input: A = [2,4]
Output: [5,4]

Note:

1 <= A.length <= 10000
1 <= B.length <= 10000
1 <= A[i] <= 100000
1 <= B[i] <= 100000
It is guaranteed that Alice and Bob have different total amounts of candy.
It is guaranteed there exists an answer.

爱丽丝和鲍勃有不同大小的糖果棒：A[i] 是爱丽丝拥有的第 i 块糖的大小，B[j] 是鲍勃拥有的第 j 块糖的大小。

因为他们是朋友，所以他们想交换一个糖果棒，这样交换后，他们都有相同的糖果总量。（一个人拥有的糖果总量是他们拥有的糖果棒大小的总和。）

返回一个整数数组 ans，其中 ans[0] 是爱丽丝必须交换的糖果棒的大小，ans[1] 是 Bob 必须交换的糖果棒的大小。

如果有多个答案，你可以返回其中任何一个。保证答案存在。

示例 1：

输入：A = [1,1], B = [2,2]
输出：[1,2]

示例 2：

输入：A = [1,2], B = [2,3]
输出：[1,2]

示例 3：

输入：A = [2], B = [1,3]
输出：[2,3]

示例 4：

输入：A = [1,2,5], B = [2,4]
输出：[5,4]

提示：

1 <= A.length <= 10000
1 <= B.length <= 10000
1 <= A[i] <= 100000
1 <= B[i] <= 100000
保证爱丽丝与鲍勃的糖果总量不同。
答案肯定存在。

500ms

 1 class Solution {
 2     func fairCandySwap(_ A: [Int], _ B: [Int]) -> [Int] {
 3         var diff = 0
 4         var i = 0
 5         var j = 0
 6         while i < A.count || j < B.count {
 7             if i < A.count {
 8                 diff -= A[i]
 9                 i += 1
10             }
11             
12             if j < B.count {
13                 diff += B[j]
14                 j += 1
15             }
16         }
17         
18         i = 0
19         j = 0
20         diff /= 2        
21         var bSet = Set<Int>()
22         for b in B {
23             bSet.insert(b)
24         }
25         
26         for a in A {
27             let pairedValue = a + diff
28             if bSet.contains(pairedValue) {
29                 return [a, pairedValue]
30             }
31         }
32         
33         return []
34     }
35 }

504ms

 1 class Solution {
 2     func fairCandySwap(_ A: [Int], _ B: [Int]) -> [Int] {
 3         let aSum = A.reduce(0) { $0 + $1 }
 4         let bSum = B.reduce(0) { $0 + $1 }
 5         let difference = bSum - aSum
 6         let bSet = Set(B.map { $0 })
 7         for a in A {
 8             if bSet.contains(a +  difference / 2) {
 9                 return [a, a +  difference / 2]
10             }
11         }
12         return [-1, -1]
13     }
14 }

Runtime: 516 ms

Memory Usage: 19.1 MB

 1 class Solution {
 2     func fairCandySwap(_ A: [Int], _ B: [Int]) -> [Int] {
 3         var dif:Int = (A.reduce(0,+) - B.reduce(0,+)) / 2
 4         var S:Set<Int> = Set<Int>()
 5         for a in A
 6         {
 7             S.insert(a)
 8         }
 9         for b in B
10         {
11             if S.contains(b + dif)
12             {
13                 return [b + dif, b]
14             }
15         }
16         return []
17     }
18 }

552ms

 1 class Solution {
 2     func fairCandySwap(_ A: [Int], _ B: [Int]) -> [Int] {
 3         let sumA = A.reduce(0, +)
 4         let sumB = B.reduce(0, +)
 5         
 6         let setA = Set(A)
 7         let setB = Set(B)
 8         
 9         for item in setA{
10             let itemB = (sumB - sumA + 2 * item) / 2
11             if setB.contains(itemB){
12                 return [item, itemB]
13             }           
14         }
15 
16         return [0,0]
17     }
18 }

556ms

 1 class Solution {
 2     func fairCandySwap(_ A: [Int], _ B: [Int]) -> [Int] {
 3         let sumA = A.reduce(0, +)
 4         let sumB = B.reduce(0, +)
 5         let bSet = Set(B)
 6         for x in A {
 7             let y = x + (sumB - sumA) / 2
 8             if bSet.contains(y){
 9                 return [x,y]
10             }            
11         }
12         return []  
13     }
14 }

580ms

 1 class Solution {
 2     func fairCandySwap(_ A: [Int], _ B: [Int]) -> [Int] {
 3         var totalA = 0
 4         var totalB = 0
 5         var mark = [Int : Bool]()
 6         for a in A {
 7             totalA += a
 8             mark[a] = true
 9         }
10         for b in B {
11             totalB += b
12         }
13         
14         let dis = (totalB - totalA) / 2
15         for b in B {
16             if let _ = mark[b - dis] {
17                 return [b - dis, b]
18             }
19         }
20         return [Int]()
21     }
22 }