★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公衆號: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: http://www.javashuo.com/article/p-gsnoupfs-kd.html
➤若是連接不是山青詠芝的博客園地址,則多是爬取做者的文章。
➤原文已修改更新!強烈建議點擊原文地址閱讀!支持做者!支持原創!
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★html
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.git
Example 1:github
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)數組
Example 2:微信
Given nums = [-2, -1, 2, 1]
, k = 1
,
return 2
. (because the subarray [-1, 2]
sums to 1 and is the longest)測試
Follow Up:
Can you do it in O(n) time?spa
給定一個數組nums和一個目標值k,找到一個子數組的最大長度總和爲k。若是沒有,則返回0。code
例1:htm
給定 nums = [1, -1, 5, -2, 3]
, k = 3
,blog
返回4。(由於子數組[1,-1,5,-2]和爲3,是最長的)
例2:
給定 nums = [-2, -1, 2, 1]
, k = 1
,
返回2。(由於子數組[-1,2]和爲1,是最長的)
跟進:
你能在O(N)時間內完成嗎?
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 }
點擊:Playground測試
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