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➤微信公众号：山青咏芝（shanqingyongzhi）➤博客园地址：山青咏芝（）➤GitHub地址：➤原文地址： ➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★Given two arrays A and B of equal size, the advantage of A with respect to B is the number of indices i for which A[i] > B[i].
Return any permutation of A that maximizes its advantage with respect to B.
Example 1:
Input: A = [2,7,11,15], B = [1,10,4,11]Output: [2,11,7,15]
Example 2:
Input: A = [12,24,8,32], B = [13,25,32,11]Output: [24,32,8,12]
Note:
1 <= A.length = B.length <= 100000 <= A[i] <= 10^90 <= B[i] <= 10^9
给定两个大小相等的数组 A 和 B，A 相对于 B 的优势可以用满足 A[i] > B[i] 的索引 i 的数目来描述。
返回 A 的任意排列，使其相对于 B 的优势最大化。
示例 1：
输入：A = [2,7,11,15], B = [1,10,4,11]输出：[2,11,7,15]
示例 2：
输入：A = [12,24,8,32], B = [13,25,32,11]输出：[24,32,8,12]
提示：
1 <= A.length = B.length <= 100000 <= A[i] <= 10^90 <= B[i] <= 10^9
Runtime: 552 ms
Memory Usage: 20.3 MB
1 class Solution { 2 func advantageCount(_ A: [Int], _ B: [Int]) -> [Int] { 3 var A = A 4 A.sort() 5 var n:Int = A.count 6 var res:[Int] = [Int](repeating:0,count:n) 7 var pq:[[Int]] = [[Int]]() 8 for i in 0.. val21 {22 res[idx] = A[hi]23 hi -= 124 }25 else26 {27 res[idx] = A[lo]28 lo += 129 }30 }31 return res32 }33 } 616ms
1 class Solution { 2 func advantageCount(_ A: [Int], _ B: [Int]) -> [Int] { 3 var leftArr = [Int]() 4 var ans = [Int](repeating: Int.min, count: A.count) 5 var tupleArrB = [(Int, Int)]() 6 for i in 0.. = sorteA.count { break }14 while aIndex < sorteA.count {15 let a = sorteA[aIndex]16 aIndex += 117 if a > v { ans[bi] = a; break}18 leftArr.append(a)19 }20 }21 22 for i in 0.. 624ms
1 class Solution { 2 func advantageCount(_ A: [Int], _ B: [Int]) -> [Int] { 3 var res = Array(repeating: 0, count: A.count) 4 5 var A = A.sorted() 6 var i = 0, e = A.count - 1 7 for (idx, n) in B.enumerated().sorted(by: { $0.1 >= $1.1 }) { 8 if A[e] > n { 9 res[idx] = A[e]10 e -= 111 } else {12 res[idx] = A[i]13 i += 114 }15 }16 return res17 }18 } 628ms
1 class Solution { 2 func advantageCount(_ A: [Int], _ B: [Int]) -> [Int] { 3 var sorted = A.sorted(by: >) 4 var sortedB = B.enumerated().sorted { 5 $0.1 > $1.1 6 } 7 var i = 0 8 var j = sorted.count - 1 9 var result = [Int](repeating: 0, count: sorted.count)10 for b in sortedB {11 if sorted[i] > b.1 {12 result[b.0] = sorted[i]13 i += 114 } else {15 result[b.0] = sorted[j]16 j -= 117 }18 }19 return result20 }21 } 772ms
1 class Solution { 2 func advantageCount(_ A: [Int], _ B: [Int]) -> [Int] { 3 var sortedA = A.sorted(by: <) 4 let sortedB = B.enumerated().map { ($0, $1) }.sorted { $0.1 < $1.1 } 5 6 var hash = [Int:Int]() 7 sortedB.forEach { (index, b) in 8 let idx = sortedA.index(where: { $0 > b }) ?? 0 9 hash[index] = sortedA.remove(at: idx)10 }11 12 return (0 ..< B.count).map { hash[$0]! }13 }14 } 976ms
1 class Solution { 2 func advantageCount(_ A: [Int], _ B: [Int]) -> [Int] { 3 let sortedA = A.sorted() 4 let sortedB = B.sorted() 5 6 var i = 0 7 var results = [Int]() 8 var wastes = [Int]() 9 10 for b in sortedB {11 while i < sortedA.count && sortedA[i] <= b {12 wastes.append(sortedA[i])13 i += 114 }15 16 if i == sortedA.count {17 results += wastes18 break19 } else {20 results.append(sortedA[i])21 i += 122 }23 }24 25 var map = [Int: [Int]]()26 27 for i in 0.. 7176ms
1 class Solution { 2 func advantageCount(_ A: [Int], _ B: [Int]) -> [Int] { 3 var sortedA = A.sorted(by: <) 4 5 var hash = [Int:Int]() 6 7 return B.map { b in 8 let idx = sortedA.index(where: { $0 > b }) ?? 0 9 return sortedA.remove(at: idx)10 }11 }12 }