[209. Minimum Size Subarray Sum](https://leetcode.com/problems/minimum-size-subarray-sum/)

### 題目描述

Given an array of positive integers `nums` and a positive integer `target`, return *the **minimal length** of a subarray whose sum is greater than or equal to* `target`. If there is no such subarray, return `0` instead.

### 範例

**Example 1:**

Input: target = 7, nums = [2,3,1,2,4,3]
Output: 2
Explanation: The subarray [4,3] has the minimal length under the problem constraint.

**Example 2:**

Input: target = 4, nums = [1,4,4]
Output: 1

**Example 3:**

Input: target = 11, nums = [1,1,1,1,1,1,1,1]
Output: 0

**Constraints**:

* 1 <= `target` <= 10^9^
* 1 <= `nums.length` <= 10^5^
* 1 <= `nums[i]` <= 10^4^

**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))`.

### 解答
#### Javascript

function minSubArrayLen(target, nums) {
  let minLen = Infinity;
  let sum = 0;
  let start = 0;
  for (let i = 0; i < nums.length; i++) {
    sum += nums[i];
    while (sum >= target) {
      const length = i - start + 1;
      minLen = Math.min(minLen, length);
      sum -= nums[start];
      start++;
    }
  }

  if (minLen === Infinity) return 0;
  return minLen;
}

> 一年多前寫過，現在寫還是錯了一次才過==
> [name=Marsgoat][time=Jul 6, 2023]

#### C#

public class Solution {
    public int MinSubArrayLen(int target, int[] nums) {
        int sum = 0;
        int min = int.MaxValue;
        int left = 0;

        for (int right = 0; right < nums.Length; right++) {
            sum += nums[right];

            while (sum >= target) {
                min = Math.Min(min, right - left + 1);
                sum = sum - nums[left++];
            }
        }

        return min == int.MaxValue ? 0 : min;
    }
}

> [name=Jim][time=Jul 6, 2023]

#### Python

class Solution:
    def minSubArrayLen(self, target: int, nums: List[int]) -> int:
        i, j, total = 0, 0, 0
        n = len(nums)
        ans = float('inf')
        while j < n:
            total += nums[j]
            j += 1
            while total >= target:
                ans = min(ans, j - i)
                total -= nums[i]
                i += 1
        return ans if ans <= n else 0

> [name=Yen-Chi Chen][time=Thu, Jul 6, 2023]

### Java

class Solution {
    public int minSubArrayLen(int target, int[] nums) {
        int windowStart = 0;
        int ans = Integer.MAX_VALUE;
        int sum = 0;

        for (int windowEnd = 0; windowEnd < nums.length; windowEnd++) {
            sum += nums[windowEnd];
            while (sum >= target) {
                ans = Math.min(ans, windowEnd - windowStart + 1);
                sum -= nums[windowStart];
                windowStart += 1;
            }
        }

        return ans != Integer.MAX_VALUE ? ans : 0;
    }
}

> [name=Ron Chen][time=Thu, Jul 6, 2023]

### Reference

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