Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous subarray of which the sum ≥ s. If there isn‘t one, return 0 instead.
Example:
Input:s = 7, nums = [2,3,1,2,4,3]Output: 2 Explanation: the subarray[4,3]has the minimal length under the problem constraint.
Follow up:
If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n).
给定一个含有 n 个正整数的数组和一个正整数 s ，找出该数组中满足其和 ≥ s 的长度最小的连续子数组。如果不存在符合条件的连续子数组，返回 0。
示例:
输入:s = 7, nums = [2,3,1,2,4,3]输出: 2 解释: 子数组[4,3]是该条件下的长度最小的连续子数组。
进阶:
如果你已经完成了O(n) 时间复杂度的解法, 请尝试 O(n log n) 时间复杂度的解法。
20ms
1 class Solution {
2 func minSubArrayLen(_ s: Int, _ nums: [Int]) -> Int {
3 guard nums.count > 0 else {
4 return 0
5 }
6 var i = 0, j = 0,minX = nums.count, sum = nums[0], match = false
7 while j < nums.count {
8 if i == j {
9 if sum < s {
10 j = j + 1
11 if j == nums.count {
12 break;
13 }
14 sum = nums[j] + sum
15 } else {
16 return 1
17 }
18 } else {
19 if sum < s {
20 j = j + 1
21 if j == nums.count {
22 break
23 }
24 sum = nums[j] + sum
25 } else {
26 match = true
27 minX = min(j - i + 1, minX)
28 sum = sum - nums[i]
29 i = i + 1
30 }
31 }
32 }
33 return match ? minX : 0
34 }
35 }
24ms
1 class Solution {
2 func minSubArrayLen(_ s: Int, _ nums: [Int]) -> Int {
3
4 var sum = 0, left = 0, res = Int.max
5 for i in nums.indices {
6 sum += nums[i]
7 while sum >= s {
8 res = min(res, i-left+1)
9 sum -= nums[left]
10 left += 1
11 }
12 }
13 return (res == Int.max) ? 0 : res
14 }
15 }
24ms
1 class Solution {
2 func minSubArrayLen(_ s: Int, _ nums: [Int]) -> Int {
3 var miniSize = Int.max ,start = 0, currentSum = 0
4
5 for (i, num) in nums.enumerated() {
6 currentSum += num
7
8 while currentSum >= s && start <= i {
9 miniSize = min(miniSize, i - start + 1)
10
11 currentSum -= nums[start]
12 start += 1
13 }
14 }
15
16 return miniSize == Int.max ? 0 : miniSize
17 }
18 }
32ms
1 class Solution {
2 func minSubArrayLen(_ s: Int, _ nums: [Int]) -> Int {
3 let n = nums.count
4 if (n < 1 || nums.reduce(0, +) < s) { return 0 }
5
6 // 维护一个滑动窗口 nums[i,j], nums[i...j] < s
7 var i = 0
8 var j = -1
9 var total = 0
10 var res = n + 1
11 while i <= n-1 {
12 if (j + 1 < n) && total < s { // 小于目标值
13 j += 1
14 total += nums[j]
15 } else { // 大于目标值
16 total -= nums[i]
17 i += 1
18 }
19 if total >= s {
20 res = min(res, j-i+1) // 求得当前最小长度
21 }
22 }
23 if res == n+1 { // 没有改变过
24 return 0
25 }
26 return res
27 }
28 }
84ms
1 class Solution {
2 func minSubArrayLen(_ s: Int, _ nums: [Int]) -> Int {
3 guard nums.count > 0 else {
4 return 0
5 }
6
7 var sums : [Int] = [0]
8 for num in nums{
9 sums.append(sums.last! + num)
10 }
11
12 var minCount : Int = Int.max
13
14 for (index, sum) in sums.enumerated(){
15 if sum >= s{
16 let key = sum - s
17 var foundIndex = binarySearch(low: 0, high: sums.count - 1, key: key, sums: sums)
18 if sums[foundIndex] > key{
19 foundIndex -= 1
20 }
21 minCount = min(minCount, index - foundIndex)
22 }
23 }
24
25 if sums.last! >= s{
26 return min(nums.count, minCount)
27 }else{
28 return 0
29 }
30
31 }
32
33 func binarySearch(low : Int, high : Int, key : Int, sums : [Int])->Int{
34 guard low < high else{
35 return low
36 }
37
38 let mid = (low + high) / 2
39 let value = sums[mid]
40 if value == key{
41 return mid
42 }else if value > key{
43 return binarySearch(low : low, high : mid-1, key : key, sums : sums)
44 }else{
45 return binarySearch(low : mid + 1, high : high, key : key, sums : sums)
46 }
47 }
48 }
原文地址：https://www.cnblogs.com/strengthen/p/10190099.html
时间： 2024-11-02 09:57:46