标签：
Implement the following operations of a queue using stacks.
Notes:
push to top, peek/pop from top, size, and is empty operations are valid.
思路：用两个stack去实现
class MyQueue { Stack<Integer> stack1=new Stack<Integer>(); //主 Stack<Integer> stack2=new Stack<Integer>(); //辅 // Push element x to the back of queue. public void push(int x) { stack1.push(x); } // Removes the element from in front of queue. public void pop() { while(!stack1.isEmpty()) { stack2.push(stack1.pop()); } if(!stack2.isEmpty()) stack2.pop(); while(!stack2.isEmpty()) { stack1.push(stack2.pop()); } } // Get the front element. public int peek() { int ret=0; while(!stack1.isEmpty()) { stack2.push(stack1.pop()); } if(!stack2.isEmpty()) { ret=stack2.peek(); } while(!stack2.isEmpty()) { stack1.push(stack2.pop()); } return ret; } // Return whether the queue is empty. public boolean empty() { return stack1.isEmpty(); } }
这里每次pop和peek操作都会将stack1转到stack2中，比较麻烦。再考虑下可以不用发给转来转去，stack1可以作为队尾，stack2可以作为对首的
class MyQueue { Stack<Integer> stack1=new Stack<Integer>(); //主 Stack<Integer> stack2=new Stack<Integer>(); //辅 // Push element x to the back of queue. public void push(int x) { stack1.push(x); } // Removes the element from in front of queue. public void pop() { if(!stack2.isEmpty()) stack2.pop(); else { while(!stack1.isEmpty()) { stack2.push(stack1.pop()); } if(!stack2.isEmpty()) stack2.pop(); } } // Get the front element. public int peek() { int ret=0; if(!stack2.isEmpty()) ret=stack2.peek(); else { while(!stack1.isEmpty()) { stack2.push(stack1.pop()); } if(!stack2.isEmpty()) ret=stack2.peek(); } return ret; } // Return whether the queue is empty. public boolean empty() { return stack1.isEmpty() && stack2.isEmpty(); } }
[LeetCode] Implement Queue using Stacks
标签：
原文地址：http://www.cnblogs.com/maydow/p/4641012.html
