基础算法：滑动窗口_python版本-开发者社区

滑动窗口算法简介

滑动窗口是一种用于处理数组或字符串子区间问题的高效算法。通过维护一个动态窗口（通常由左右指针定义），在遍历过程中调整窗口大小或位置，避免重复计算，将时间复杂度从O(n²)优化至O(n)。适用于连续子数组、子串等问题。

滑动窗口的典型应用场景

寻找满足条件的最小/最大子数组长度（如和≥target的最短子数组）。
统计满足特定条件的子串或子数组数量（如包含所有字符的最短覆盖子串）。
处理固定长度窗口的问题（如计算大小为k的子数组平均值）。

Python实现模板

可变窗口大小（通用模板）

def sliding_window(s, target): left = 0 result = 0 # 或其他初始值（如float('inf')） current_value = 0 # 当前窗口的统计值（如和、字符频率等） for right in range(len(s)): current_value += s[right] # 更新右指针带来的影响 while 窗口满足条件（如current_value >= target）: result = min(result, right - left + 1) # 更新结果（根据问题调整） current_value -= s[left] # 移除左指针的影响 left += 1 # 收缩窗口 return result

固定窗口大小（示例：大小为k的子数组平均值）

def fixed_window(nums, k): window_sum = sum(nums[:k]) result = [window_sum / k] for i in range(k, len(nums)): window_sum += nums[i] - nums[i - k] # 滑动窗口：移除左端，加入右端 result.append(window_sum / k) return result

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实战案例

案例1：和≥target的最短子数组（LeetCode 209）

def minSubArrayLen(target, nums): left = 0 min_len = float('inf') current_sum = 0 for right in range(len(nums)): current_sum += nums[right] while current_sum >= target: min_len = min(min_len, right - left + 1) current_sum -= nums[left] left += 1 return min_len if min_len != float('inf') else 0

案例2：无重复字符的最长子串（LeetCode 3）

def lengthOfLongestSubstring(s): left = 0 max_len = 0 char_index = {} # 记录字符最后出现的位置 for right in range(len(s)): if s[right] in char_index and char_index[s[right]] >= left: left = char_index[s[right]] + 1 # 移动左指针到重复字符的下一位 char_index[s[right]] = right max_len = max(max_len, right - left + 1) return max_len