[Swift]LeetCode325. 最大子数组之和为k $ Maximum Size Subarray Sum Equals k

Given an array nums and a target value k, find the maximum length of a subarray that sums to k. If there isn't one, return 0 instead.

Example 1:

Given nums = [1, -1, 5, -2, 3]k = 3,
return 4. (because the subarray [1, -1, 5, -2] sums to 3 and is the longest)

Example 2:

Given nums = [-2, -1, 2, 1]k = 1,
return 2. (because the subarray [-1, 2] sums to 1 and is the longest)

Follow Up:
Can you do it in O(n) time?


给定一个数组nums和一个目标值k,找到一个子数组的最大长度总和为k。如果没有,则返回0。

例1:

给定 nums = [1, -1, 5, -2, 3]k = 3,

返回4。(因为子数组[1,-1,5,-2]和为3,是最长的)

例2:

给定 nums = [-2, -1, 2, 1]k = 1,

返回2。(因为子数组[-1,2]和为1,是最长的)

跟进:

你能在O(N)时间内完成吗?


Solution:

 1 class Solution {
 2     func maxSubArrayLen(_ nums:inout [Int],_ k:Int) -> Int {
 3         var sum:Int = 0
 4         var res:Int = 0
 5         var m:[Int:Int] = [Int:Int]()
 6         for i in 0..<nums.count
 7         {
 8             sum += nums[i]
 9             if sum == k
10             {
11                 res = i + 1
12             }
13             else if m[sum - k] != nil
14             {
15                 res = max(res, i - m[sum - k,default:0])
16             }
17             if m[sum] == nil
18             {
19                 m[sum] = i
20             }
21         }
22         return res        
23     }
24 }

点击:Playground测试

1 let k1:Int = 3
2 var nums1:[Int] = [1, -1, 5, -2, 3]
3 print(Solution().maxSubArrayLen(&nums1,k1))
4 //Print 4
5 
6 let k2:Int = 1
7 var nums2:[Int] = [-2, -1, 2, 1]
8 print(Solution().maxSubArrayLen(&nums2,k2))
9 //Print 2

 

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