[Swift]LeetCode1131. 绝对值表达式的最大值 | Maximum of Absolute Value Expression
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Given two arrays of integers with equal lengths, return the maximum value of:
|arr1[i] - arr1[j]| + |arr2[i] - arr2[j]| + |i - j|
where the maximum is taken over all 0 <= i, j < arr1.length
.
Example 1:
Input: arr1 = [1,2,3,4], arr2 = [-1,4,5,6] Output: 13
Example 2:
Input: arr1 = [1,-2,-5,0,10], arr2 = [0,-2,-1,-7,-4] Output: 20
Constraints:
2 <= arr1.length == arr2.length <= 40000
-10^6 <= arr1[i], arr2[i] <= 10^6
给你两个长度相等的整数数组,返回下面表达式的最大值:
|arr1[i] - arr1[j]| + |arr2[i] - arr2[j]| + |i - j|
其中下标 i
,j
满足 0 <= i, j < arr1.length
。
示例 1:
输入:arr1 = [1,2,3,4], arr2 = [-1,4,5,6] 输出:13
示例 2:
输入:arr1 = [1,-2,-5,0,10], arr2 = [0,-2,-1,-7,-4] 输出:20
提示:
2 <= arr1.length == arr2.length <= 40000
-10^6 <= arr1[i], arr2[i] <= 10^6
192ms
1 class Solution { 2 func maxAbsValExpr(_ arr1: [Int], _ arr2: [Int]) -> Int { 3 let n = arr1.count 4 var closet = 0 5 var cur = 0 6 var res = 0 7 for p in [-1,1] { 8 for q in [-1, 1] { 9 closet = arr1[0] * p + arr2[0] * q 10 for i in 1 ..< n { 11 cur = arr1[i] * p + arr2[i] * q + i 12 res = max(res, cur - closet) 13 closet = min(closet, cur) 14 } 15 } 16 } 17 return res 18 } 19 }
Runtime: 196 ms
Memory Usage: 21.3 MB
1 class Solution { 2 func maxAbsValExpr(_ arr1: [Int], _ arr2: [Int]) -> Int { 3 var n:Int = arr1.count 4 var ans:Int = 0 5 var maxmm:Int = Int.min / 2 6 var maxmp:Int = Int.min / 2 7 var maxpm:Int = Int.min / 2 8 var maxpp:Int = Int.min / 2 9 for i in 0..<n 10 { 11 maxmm = max(maxmm, -arr1[i] - arr2[i]-i) 12 maxmp = max(maxmp, -arr1[i] + arr2[i]-i) 13 maxpm = max(maxpm, arr1[i] - arr2[i]-i) 14 maxpp = max(maxpp, arr1[i] + arr2[i]-i) 15 ans = max(ans, maxmm + arr1[i] + arr2[i]+i) 16 ans = max(ans, maxmp + arr1[i] - arr2[i]+i) 17 ans = max(ans, maxpm - arr1[i] + arr2[i]+i) 18 ans = max(ans, maxpp - arr1[i] - arr2[i]+i) 19 } 20 return ans 21 } 22 }
204ms
1 class Solution { 2 func maxAbsValExpr(_ arr1: [Int], _ arr2: [Int]) -> Int { 3 var ans = 0 4 5 for i in [-1,1] { 6 for j in [-1,1]{ 7 var best = arr1[0] * i + arr2[0] * j 8 for k in 1..<arr1.count { 9 ans = max(ans, arr1[k]*i + arr2[k]*j + k - best) 10 best = min(best, arr1[k]*i + arr2[k]*j + k) 11 } 12 } 13 } 14 return ans 15 } 16 }
224ms
1 class Solution { 2 func maxAbsValExpr(_ arr1: [Int], _ arr2: [Int]) -> Int { 3 var res = 0 4 var maxmm = Int.min 5 var maxmp = Int.min 6 var maxpm = Int.min 7 var maxpp = Int.min 8 9 var i = 0 10 while i < arr1.count { 11 maxmm = max(maxmm, -arr1[i] - arr2[i] - i) 12 maxmp = max(maxmp, -arr1[i] + arr2[i] - i) 13 maxpm = max(maxpm, arr1[i] - arr2[i] - i) 14 maxpp = max(maxpp, arr1[i] + arr2[i] - i) 15 res = max(res, maxmm + arr1[i] + arr2[i] + i) 16 res = max(res, maxmp + arr1[i] - arr2[i] + i) 17 res = max(res, maxpm - arr1[i] + arr2[i] + i) 18 res = max(res, maxpp - arr1[i] - arr2[i] + i) 19 20 i += 1 21 } 22 23 return res; 24 } 25 }
252ms
1 class Solution { 2 func maxAbsValExpr(_ arr1: [Int], _ arr2: [Int]) -> Int { 3 var aux = [Int](repeating: 0, count: arr1.count) 4 var result = Int.min 5 for j in [-1, 1] { 6 for k in [-1, 1] { 7 8 for i in stride(from: 0, to: arr1.count, by: 1) { 9 aux[i] = arr1[i] * j + arr2[i] * k + i 10 } 11 let iPart = aux.max()! 12 let jPart = aux.min()! 13 result = max(result, iPart - jPart) 14 } 15 } 16 return result 17 } 18 }