归档 2015年1月

Implement the reversal of a singly linked list iteratively and recursively.

链表逆置既可以用迭代实现也可以使用递归实现。在我看来，迭代法应该比其递归形式的等价实现更加高效，内存开销更小。（试想对一个包含百万个元素的链表执行递归的逆置操作!很快栈空间就会用尽）

递归实现的代码行数更少，但是要想把代码写对会比较困难。另一方面，迭代实现代码行数较多但是比较容易验证。

1) 迭代实现:

void reverse(Node*& head) {
  if (!head) return;
  Node* prev = NULL;
  Node* curr = head;
  while (curr) {
    Node ...

题目描述：

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,

return 1->4->3->2->5->NULL.

Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ...

题目描述：

Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

You must do this in-place without altering the nodes' values.

For example,
Given {1,2,3,4}, reorder it to {1,4,2,3}.

题目大意：

给定一个单链表L：L0→L1→…→Ln-1→Ln
重新将其排序为 ...

题目描述：

Given a binary tree, return the preorder traversal of its nodes' values.

For example:

Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

return [1,2,3].

题目大意：

给定一棵二叉树，返回节点值的先序遍历结果（尽量使用非递归算法）。

解题思路：

使用数据结构栈（Stack）

先序遍历非递归算法的伪代码如下（摘自Wikipedia）：

iterativePreorder(node)
  parentStack ...

题目描述：

You are a traveling salesman. You have already heard a lot about how hard the problems of a traveling salesman can be. Luckily, the one you currently have seems easier.

There are M cities where you can sell products ...