算法设计与应用基础-第六周

Remove Duplicates from Sorted List          

Given a sorted linked list, delete all duplicates such that each element appear onlyonce.

For example,

Given

1->1->2

, return

1->2

.

Given

1->1->2->3->3

, return

1->2->3

.

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* deleteDuplicates(ListNode* head)
{
if(head==NULL)
return head;
else if(head!=NULL)
{
ListNode* pre=head;
ListNode* nex=head->next;
while(nex)
{
if(pre->val==nex->val)
{
pre->next=nex->next;
nex=nex->next;
}
else if(pre->val!=nex->val)
{
pre=pre->next;
nex=nex->next;
}
}
return head;
}
}
};

Binary Tree Level Order Traversal II          

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:

Given binary tree

[3,9,20,null,null,15,7]

,

3
/ \
9  20
/  \
15   7

return its bottom-up level order traversal as:

[
[15,7],
[9,20],
[3]
]

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int>> levelOrderBottom(TreeNode* root)
{
vector<vector<int>> res;
if (root == NULL)
return res;
queue<TreeNode *> q;
q.push(root);
while (!q.empty())
{
int l = q.size();
vector<int> node;
for (int i = 0; i < l; i++)
{
TreeNode *cur = q.front();
q.pop();
node.push_back(cur->val);
if (cur->left != NULL) q.push(cur->left);
if (cur->right != NULL) q.push(cur->right);
}
res.push_back(node);
}
reverse(res.begin(), res.end());
return res;
}

};