[Swift]LeetCode209. 长度最小的子数组 | Minimum Size Subarray Sum

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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: 

s = 7, nums = [2,3,1,2,4,3]
[4,3]

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: 

s = 7, nums = [2,3,1,2,4,3]
[4,3]

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

class Solution {
func minSubArrayLen(_ s: Int, _ nums: [Int]) -> Int {
    guard nums.count > 0 else {
        return 0
    }
    var i = 0, j = 0,minX = nums.count, sum = nums[0], match = false
    while j < nums.count {
        if i == j {
            if sum < s {
                j = j + 1
                if j == nums.count {
                    break;
                }
                sum = nums[j] + sum
            } else {
                return 1
            }
        } else {
            if sum < s {
                j = j + 1
                if j == nums.count {
                    break
                }
                sum = nums[j] + sum
            } else {
                match = true
                minX = min(j - i + 1, minX)
                sum = sum - nums[i]
                i = i + 1
            }
        }
    }
    return match ? minX : 0
}
}

---

24ms

class Solution {
    func minSubArrayLen(_ s: Int, _ nums: [Int]) -> Int {

        var sum = 0, left = 0, res = Int.max
        for i in nums.indices {
            sum += nums[i]
            while sum >= s {
                res = min(res, i-left+1)
                sum -= nums[left]
                left += 1
            }
        }
        return (res == Int.max) ? 0 : res
    }
}

---

24ms

class Solution {
    func minSubArrayLen(_ s: Int, _ nums: [Int]) -> Int {
        var miniSize = Int.max ,start = 0, currentSum = 0
        
        for (i, num) in nums.enumerated() {
            currentSum += num
            
            while currentSum >= s && start <= i {
                miniSize = min(miniSize, i - start + 1)
                
                currentSum -= nums[start]
                start += 1
            }
        }
        
        return miniSize == Int.max ? 0 : miniSize
    }
}

---

32ms

class Solution {
    func minSubArrayLen(_ s: Int, _ nums: [Int]) -> Int {
        let n = nums.count
        if (n < 1 || nums.reduce(0, +) < s) { return 0 }

        // 维护一个滑动窗口 nums[i,j], nums[i...j] < s
        var i = 0
        var j = -1
        var total = 0
        var res = n + 1
        while i <= n-1 {
            if (j + 1 < n) && total < s {   // 小于目标值
                j += 1
                total += nums[j]
            } else {    // 大于目标值
                total -= nums[i]
                i += 1
            }
            if total >= s {
                res = min(res, j-i+1)   // 求得当前最小长度
            }
        }
        if res == n+1 { // 没有改变过
            return 0
        }
        return res
    }
}

---

84ms

class Solution {
    func minSubArrayLen(_ s: Int, _ nums: [Int]) -> Int {
        guard nums.count > 0 else {
            return 0
        }
        
        var sums : [Int] = [0]
        for num in nums{
            sums.append(sums.last! + num)
        }
        
        var minCount : Int = Int.max
        
        for (index, sum) in sums.enumerated(){
            if sum >= s{
                let key = sum - s
                var foundIndex = binarySearch(low: 0, high: sums.count - 1, key: key, sums: sums)
                if sums[foundIndex] > key{
                    foundIndex -= 1
                }
                minCount = min(minCount, index - foundIndex)
            }
        }
        
         if sums.last! >= s{
            return min(nums.count, minCount)
        }else{
            return 0 
        }
        
    }
    
    func binarySearch(low : Int, high : Int, key : Int, sums : [Int])->Int{
        guard low < high else{
            return low
        }
        
        let mid = (low + high) / 2
        let value = sums[mid]
        if value == key{
            return mid
        }else if value > key{
            return binarySearch(low : low, high : mid-1, key : key, sums : sums)
        }else{
            return binarySearch(low : mid + 1, high : high, key : key, sums : sums)
        }
    }
}