比较简单的几道题，不做详细的解释，只为之后可以回头看看自己之前的代码!!

虽然几道题都比较简单，但感觉自己写得不好，希望博文如果被哪位大神看到，可以留言下写下你的做法！

Reverse Linked List II

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 ≤ length of list.

AC代码：

public class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
		if (head == null)
			return null;
		if (m == n)
        	return head;
        ListNode pHead = new ListNode(0);
        pHead.next = head;
        ListNode preMNode = getListNode(pHead, m-1);
        ListNode MNode = getListNode(pHead, m);
        ListNode preNode = preMNode;
        
        int len = n-m;
        for (int i=0; i<=len; ++i){
        	ListNode MNodeNext = MNode.next;
        	MNode.next = preNode;
        	preNode = MNode;
        	MNode = MNodeNext;
        }
        
        preMNode.next.next = MNode;
        preMNode.next = preNode;
        
		return pHead.next;
    }
	public ListNode getListNode(ListNode startNode, int index){
		for (int i=0; i<index; ++i){
			startNode = startNode.next;
		}
		return startNode;
	}
	
	
}

Binary Tree Inorder Traversal

Given a binary tree, return the inorder traversal of its nodes‘ values.

For example:
Given binary tree {1,#,2,3},

   1
         2
    /
   3

return [1,3,2].

Note: Recursive solution is trivial, could you do it iteratively?

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

public class Solution {
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        ArrayList<Integer> arrays = new ArrayList<Integer>();
        if (root == null)
            return arrays;
       
        
        Stack<TreeNode> stack = new Stack<TreeNode>();
        stack.push(root);
        
        while (!stack.empty()){
            TreeNode topNode = stack.peek();
            if (topNode.left != null){
                stack.push(topNode.left);
                topNode.left = null;
                continue;
            }
            
            stack.pop();
            arrays.add(topNode.val);
            
            if (topNode.right != null){
                stack.push(topNode.right);
            }
            
        }
        return arrays;
    }
}

Merge Two Sorted Lists (关于两个已经排好序的链表的合并)

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

AC代码：

public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
		ListNode list = new ListNode(0);
		ListNode newlist = list;
		if (l1 == null){
			list = l2;
			return list;
		}
		if (l2 == null){
			list = l1;
			return list;
		}
		while (l1 != null && l2 != null){
			if (l1.val < l2.val){
				list.next = l1;
				list = list.next;
				l1 = l1.next;
			}else{
				list.next = l2;
				list = list.next;
				l2 = l2.next;
			}
		}
		if (l1 != null){
			list.next = l1;
		}
		if (l2 != null){
			list.next = l2;
		}
        return newlist.next;
    }
}

Same Tree

Given two binary trees, write a function to check if they are equal or not.

Two binary trees are considered equal if they are structurally identical and the nodes have the same value.

AC代码：

public class Solution {
    public boolean isSameTree(TreeNode p, TreeNode q) {
        if (p == null && q == null)
            return true;
        if (p == null)
            return false;
        if (q == null)
            return false;
        
        if (p.val != q.val)
            return false;
        return isSameTree(p.left,q.left) && isSameTree(p.right,q.right);
    }
}

leetCode解题报告之几道基础题,布布扣,bubuko.com

leetCode解题报告之几道基础题

原文：http://blog.csdn.net/ljphhj/article/details/21865025