网上的若干算法都太复杂了，现提出包氏算法如下：

先for循环把arr1中的元素入栈并在每次遍历时检索arr2中可以pop的元素。如果循环结束而stack中还有元素就说明arr2序列不是pop序列。static bool JudgeSequenceIsPossible(int[] arr1, int[] arr2){Stack stack new Stack();for (int i 0, j 0; i arr1.Length; i){stack.Push(arr1[i]);while (stack.Count 0 (int)stack.Peek() arr2[j]){stack.Pop();j;}}return stack.Count 0;}6.递归反转一个栈要求不得重新申请一个同样的栈空间复杂度o(1)算法思想汉诺塔的思想非常复杂玩过九连环的人都想得通的static void ReverseStack(ref Stack stack) { if (stack.Count 0) return; object top stack.Pop(); ReverseStack(ref stack); if (stack.Count 0) { stack.Push(top); return; } object top2 stack.Pop(); ReverseStack(ref stack); stack.Push(top); ReverseStack(ref stack); stack.Push(top2); }7.给栈排个序本题目是上一题目的延伸static void Sort(ref Stack stack) { if (stack.Count 0) return; object top stack.Pop(); Sort(ref stack); if (stack.Count 0) { stack.Push(top); return; } object top2 stack.Pop(); if ((int)top (int)top2) { stack.Push(top); Sort(ref stack); stack.Push(top2); } else { stack.Push(top2); Sort(ref stack); stack.Push(top); } }8..如何用一个数组实现两个栈继续我所提倡的抠门儿思想也不枉我和青菜脸相交一场。网上流传着两种方法方法1 采用交叉索引的方法一号栈所占数组索引为0, 2, 4, 6, 8......(K*2)二号栈所占数组索引为13579 ......(K*2 1)算法实现如下public class NewStack { object[] arr; int top1; int top2; public NewStack(int capticy) { arr new object[capticy]; top1 -1; top2 -2; } public void Push(int type, object element) { if (type 1) { if (top1 2 arr.Length) throw new Exception(The stack is full); else { top1 2; arr[top1] element; } } else //type2 { if (top2 2 arr.Length) throw new Exception(The stack is full); else { top2 2; arr[top2] element; } } } public object Pop(int type) { object obj null; if (type 1) { if (top1 -1) throw new Exception(The stack is empty); else { obj arr[top1]; arr[top1] null; top1 - 2; } } else //type 2 { if (top2 -2) throw new Exception(The stack is empty); else { obj arr[top2]; arr[top2] null; top2 - 2; } } return obj; } public object Peek(int type) { if (type 1) { if (top1 -1) throw new Exception(The stack is empty); return arr[top1]; } else //type 2 { if (top2 -2) throw new Exception(The stack is empty); return arr[top2]; } } }方法2第一个栈A从最左向右增长第二个栈B从最右向左增长代码实现如下public class NewStack { object[] arr; int top1; int top2; public NewStack(int capticy) { arr new object[capticy]; top1 0; top2 capticy; } public void Push(int type, object element) { if (top1 top2) throw new Exception(The stack is full); if (type 1) { arr[top1] element; top1; } else //type2 { top2--; arr[top2] element; } } public object Pop(int type) { object obj null; if (type 1) { if (top1 0) throw new Exception(The stack is empty); else { top1--; obj arr[top1]; arr[top1] null; } } else //type 2 { if (top2 arr.Length) throw new Exception(The stack is empty); else