力扣题解最佳实践  >  数组  >  209. 长度最小的子数组  >  已用 Java, Python, JavaScript, C#, Go, Ruby, C++ 语言实现  >  转发 或 贡献代码

    力扣链接：209. 长度最小的子数组，难度：中等。

    给定一个含有 n 个正整数的数组和一个正整数 target 。

    找出该数组中满足其总和大于等于 target 的长度最小的 子数组 [numsl, numsl+1, ..., numsr-1, numsr] ，并返回其长度。如果不存在符合条件的子数组，返回 0 。

    子数组 是数组中连续的 非空 元素序列。

    示例 1:

    输入: target = 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子数组。

    示例 2:

    输入: target = 4, nums = [1,4,4]

    输出: 1

    解释: target = 11, nums = [1,1,1,1,1,1,1,1]

    示例 3:

    输入: target = 11, nums = [1,1,1,1,1,1,1,1]

    输出: 0

    约束:

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

    思路
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    1. 对于子数组问题，可以考虑使用滑动窗口技术，它类似于快慢双指针方法。

    步骤

    1. 遍历 nums 数组，元素的 index 可命名为 fastIndex。虽然不起眼，但这是 快慢指针技术 最重要的逻辑。请最好记住它。

    2. sum += nums[fast_index].

      var minLength = Integer.MAX_VALUE;
var sum = 0;
var slowIndex = 0;

for (var fastIndex = 0; fastIndex < nums.length; fastIndex++) { // 本行是`快慢指针技术`最重要的逻辑
    sum += nums[fastIndex]; // 1
}

return minLength;

    3. 控制slowIndex。

      var minLength = Integer.MAX_VALUE;
var sum = 0;
var slowIndex = 0;

for (var fastIndex = 0; fastIndex < nums.length; fastIndex++) {
    sum += nums[fastIndex];

    while (sum >= target) { // 1
        minLength = Math.min(minLength, fastIndex - slowIndex + 1); // 2
        sum -= nums[slowIndex]; // 3
        slowIndex++; // 4
    }
}

if (minLength == Integer.MAX_VALUE) { // 5
    return 0; // 6
}

return minLength;

    复杂度

    时间复杂度

    O(N)

    空间复杂度

    O(1)

    题解语言： Java Python JavaScript C# Go Ruby C++

    Java # 

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

        for (var fastIndex = 0; fastIndex < nums.length; fastIndex++) { // This line is the most important. You'd better memorize it.
            sum += nums[fastIndex];

            while (sum >= target) {
                minLength = Math.min(minLength, fastIndex - slowIndex + 1);
                sum -= nums[slowIndex];
                slowIndex++;
            }
        }

        if (minLength == Integer.MAX_VALUE) {
            return 0;
        }

        return minLength;
    }
}

    Python # 

    class Solution:
    def minSubArrayLen(self, target: int, nums: List[int]) -> int:
        min_length = float('inf')
        sum_ = 0
        slow_index = 0

        for fast_index, num in enumerate(nums): # This line is the most important. You'd better memorize it.
            sum_ += num

            while sum_ >= target:
                min_length = min(min_length, fast_index - slow_index + 1)
                sum_ -= nums[slow_index]
                slow_index += 1

        if min_length == float('inf'):
            return 0

        return min_length

    JavaScript # 

    var minSubArrayLen = function (target, nums) {
  let minLength = Number.MAX_SAFE_INTEGER
  let sum = 0
  let slowIndex = 0

  nums.forEach((num, fastIndex) => { // This line is the most important. You'd better memorize it.
    sum += num

    while (sum >= target) {
      minLength = Math.min(minLength, fastIndex - slowIndex + 1)
      sum -= nums[slowIndex]
      slowIndex++
    }
  })

  if (minLength == Number.MAX_SAFE_INTEGER) {
    return 0
  }

  return minLength
};

    C# # 

    public class Solution
{
    public int MinSubArrayLen(int target, int[] nums)
    {
        int minLength = Int32.MaxValue;
        int sum = 0;
        int slowIndex = 0;

        for (int fastIndex = 0; fastIndex < nums.Length; fastIndex++) // This line is the most important. You'd better memorize it.
        {
            sum += nums[fastIndex];

            while (sum >= target)
            {
                minLength = Math.Min(minLength, fastIndex - slowIndex + 1);
                sum -= nums[slowIndex];
                slowIndex++;
            }
        }

        if (minLength == Int32.MaxValue)
            return 0;

        return minLength;
    }
}

    Go # 

    func minSubArrayLen(target int, nums []int) int {
    minLength := math.MaxInt32
    sum := 0
    slowIndex := 0

    for fastIndex := 0; fastIndex < len(nums); fastIndex++ { // This line is the most important. You'd better memorize it.
        sum += nums[fastIndex]

        for sum >= target {
            minLength = min(minLength, fastIndex - slowIndex + 1)
            sum -= nums[slowIndex]
            slowIndex++
        }
    }

    if minLength == math.MaxInt32 {
        return 0
    }

    return minLength
}

func min(a, b int) int {
    if a < b {
        return a
    }
    return b
}

    Ruby # 

    # @param {Integer} target
# @param {Integer[]} nums
# @return {Integer}
def min_sub_array_len(target, nums)
  min_length = Float::INFINITY
  sum = 0
  slow_index = 0

  nums.each_with_index do |num, fast_index| # This line is the most important. You'd better memorize it.
    sum += num

    while sum >= target
      min_length = [min_length, fast_index - slow_index + 1].min
      sum -= nums[slow_index]
      slow_index += 1
    end
  end

  min_length == Float::INFINITY ? 0 : min_length
end

    C++ # 

    class Solution {
public:
    int minSubArrayLen(int target, vector<int>& nums) {
        int min_length = INT_MAX;
        int sum = 0;
        int slow_index = 0;

        for (int fast_index = 0; fast_index < nums.size(); fast_index++) {
            sum += nums[fast_index];

            while (sum >= target) {
                min_length = min(min_length, fast_index - slow_index + 1);
                sum -= nums[slow_index];
                slow_index++;
            }
        }

        if (min_length == INT_MAX) {
            return 0;
        }

        return min_length;
    }
};
    题解语言： Java Python JavaScript C# Go Ruby C++

    其它语言

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