### 方法一：排序后交替插入 ```python def sort_kids_and_distance_1(heights): sorted_heights = sorted(heights) result = [] left, right = 0, len(sorted_heights) - 1 while left <= right: if left == right: result.append(sorted_heights[left]) break result.append(sorted_heights[left]) result.append(sorted_heights[right]) left += 1 right -= 1 min_distance = 0 for i in range(1, len(result)): min_distance += abs(result[i] - result[i - 1]) return min_distance ``` ### 方法二：奇偶位置分别填充 ```python def sort_kids_and_distance_2(heights): sorted_heights = sorted(heights) result = [0] * len(heights) odd_index = 0 even_index = len(heights) - 1 for i in range(len(heights)): if i % 2 == 0: result[i] = sorted_heights[odd_index] odd_index += 1 else: result[i] = sorted_heights[even_index] even_index -= 1 min_distance = 0 for i in range(1, len(result)): min_distance += abs(result[i] - result[i - 1]) return min_distance ``` ### 方法三：使用双指针和列表推导式 ```python def sort_kids_and_distance_3(heights): sorted_heights = sorted(heights) left = 0 right = len(sorted_heights) - 1 result = [sorted_heights[left] if i % 2 == 0 else sorted_heights[right] for i in range(len(heights))] min_distance = 0 for i in range(1, len(result)): min_distance += abs(result[i] - result[i - 1]) return min_distance ``` ### 方法四：分奇偶列表再合并 ```python def sort_kids_and_distance_4(heights): sorted_heights = sorted(heights) odd_heights = sorted_heights[:len(heights) // 2] even_heights = sorted_heights[len(heights) // 2:][::-1] result = [] for i in range(len(odd_heights)): result.extend([odd_heights[i], even_heights[i]]) if len(heights) % 2 == 1: result.append(even_heights[-1]) min_distance = 0 for i in range(1, len(result)): min_distance += abs(result[i] - result[i - 1]) return min_distance ``` ### 方法五：使用堆排序思想 ```python import heapq def sort_kids_and_distance_5(heights): min_heap = heights.copy() heapq.heapify(min_heap) max_heap = [-x for x in heights] heapq.heapify(max_heap) result = [] while min_heap and max_heap: if len(result) % 2 == 0: result.append(heapq.heappop(min_heap)) else: result.append(-heapq.heappop(max_heap)) if min_heap: result.append(heapq.heappop(min_heap)) min_distance = 0 for i in range(1, len(result)): min_distance += abs(result[i] - result[i - 1]) return min_distance ``` 以下是测试代码： ```python heights = [1, 2, 3, 4, 5] print(sort_kids_and_distance_1(heights)) print(sort_kids_and_distance_2(heights)) print(sort_kids_and_distance_3(heights)) print(sort_kids_and_distance_4(heights)) print(sort_kids_and_distance_5(heights)) ```