2017/11/21 Leetcode 日记
2017/11/21 Leetcode 日记
496. Next Greater Element I
You are given two arrays (without duplicates) nums1 and nums2 where nums1’s elements are subset of nums2. Find all the next greater numbers for nums1's elements in the corresponding places of nums2.
The Next Greater Number of a number x in nums1 is the first greater number to its right in nums2. If it does not exist, output -1 for this number.
class Solution {
public:
vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
vector<int> tNums;
for(int i = , sz = findNums.size(); i < sz; i++){
bool finded = false;
int index = ;
for(int j = findN(findNums[i], nums), nsz = nums.size(); j < nsz; j++){
if(nums[j] > findNums[i]){
index = j;
break;
}
}
if(index == ) tNums.push_back(-);
else tNums.push_back(nums[index]);
}
return tNums;
}
// return index of nums[k] == num
int findN(int num, vector<int>& nums){
for(int i = , sz = nums.size(); i < sz; i++){
if(nums[i] == num){
return i;
}
}
return -;
}
};
c++
513. Find Bottom Left Tree Value
Given a binary tree, find the leftmost value in the last row of the tree.
(BFS)
/**
* 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:
int findBottomLeftValue(TreeNode* root) {
queue<TreeNode*> leaves;
leaves.push(root);
while(!leaves.empty()){
TreeNode *temp = leaves.front();leaves.pop();
if(!temp->right && !temp->left && leaves.empty()) return temp->val;
if(temp->right)
leaves.push(temp->right);
if(temp->left)
leaves.push(temp->left);
}
}
};
c++
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None class Solution:
def findBottomLeftValue(self, root):
"""
:type root: TreeNode
:rtype: int
"""
list = []
list.append(root)
while(len(list)):
temp = list.pop()
if(temp.right == None and temp.left == None and len(list) == ):
return temp.val
if(temp.right):
list.append(temp.right)
if(temp.left):
list.append(temp.left)
python3
540. Single Element in a Sorted Array
Given a sorted array consisting of only integers where every element appears twice except for one element which appears once. Find this single element that appears only once.
(二分搜索)
class Solution {
public:
int singleNonDuplicate(vector<int>& nums) {
int left = , right = nums.size()-;
int mid = (left + right) / ;
if(mid == left) return nums[mid];
while(left < right){
if(mid % == ){
if(Left(mid, nums)) {
right = mid-;
mid = (right + left)/;
}
else if(Right(mid, nums)){
left = mid+;
mid = (left + right)/;
}
else return nums[mid];
}else{
if(Left(mid, nums)){
left = mid+;
mid = (left + right)/;
}else{
right = mid-;
mid = (right + left)/;
}
}
}
return nums[mid];
}
bool Left(int i, vector<int>& nums){
int left = ;
if(i == ) return false;
else if (nums[i] != nums[i-]) return false;
else return true;
}
bool Right(int i, vector<int>& nums){
int right = nums.size()-;
if (i == right) return false;
else if (nums[i] != nums[i+]) return false;
else return true;
}
};
c++
647. Palindromic Substrings
Given a string, your task is to count how many palindromic substrings in this string.
The substrings with different start indexes or end indexes are counted as different substrings even they consist of same characters.
class Solution {
public:
int countSubstrings(string s) {
int left = , right = s.size();
// cout<<right<<endl;
// return traceBack(s, left, right);
int count = ;
for(int i = left; i < right; i++){
for(int j = i; j < right; j++){
bool palindromic = true;
for(int ind = i, end = j; ind <= end; ind++, end--){
if(s[ind] != s[end]) palindromic = false;
}
if(palindromic) count++;
}
}
return count;
}
};
c++
637. Average of Levels in Binary Tree
Given a non-empty binary tree, return the average value of the nodes on each level in the form of an array.
/**
* 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<double> averageOfLevels(TreeNode* root) {
queue<TreeNode*> q;
queue<long long> level;
vector<double> ave; q.push(root);
level.push();
long long last = , sum = , num = ;
while(!q.empty()){
TreeNode * temp = q.front();q.pop();
long long tp = level.front(); level.pop(); if(temp->right) {q.push(temp->right);level.push(tp+);}
if(temp->left) {q.push(temp->left);level.push(tp+);} if(tp == last){
sum += temp->val;
num ++;
}else{
ave.push_back((double)sum/(double)num);
sum = ;
num = ;
last = tp;
sum += temp->val;
}
if(q.empty()) ave.push_back((double)sum/(double)num);
}
return ave;
}
};
c++
515. Find Largest Value in Each Tree Row
You need to find the largest value in each row of a binary tree.
/**
* 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<int> largestValues(TreeNode* root) {
queue<TreeNode*> q;
queue<int> level;
vector<int> ave; if(root == NULL) return ave; q.push(root);
level.push();
int last = , max = -(<<);
while(!q.empty()){
TreeNode * temp = q.front();q.pop();
int tp = level.front(); level.pop(); if(temp->right) {q.push(temp->right);level.push(tp+);}
if(temp->left) {q.push(temp->left);level.push(tp+);} if(tp == last){
if(max < temp->val)
max = temp->val;
}else{
ave.push_back(max);
max = -(<<);
last = tp;
if(max < temp->val)
max = temp->val;
}
if(q.empty()) ave.push_back(max);
}
return ave;
}
};
c++
2017/11/21 Leetcode 日记的更多相关文章
- 2017/11/22 Leetcode 日记
2017/11/22 Leetcode 日记 136. Single Number Given an array of integers, every element appears twice ex ...
- 2017/11/13 Leetcode 日记
2017/11/13 Leetcode 日记 463. Island Perimeter You are given a map in form of a two-dimensional intege ...
- 2017/11/20 Leetcode 日记
2017/11/14 Leetcode 日记 442. Find All Duplicates in an Array Given an array of integers, 1 ≤ a[i] ≤ n ...
- 2017/11/9 Leetcode 日记
2017/11/9 Leetcode 日记 566. Reshape the Matrix In MATLAB, there is a very useful function called 'res ...
- 2017/11/7 Leetcode 日记
2017/11/7 Leetcode 日记 669. Trim a Binary Search Tree Given a binary search tree and the lowest and h ...
- 2017/11/6 Leetcode 日记
2017/11/6 Leetcode 日记 344. Reverse String Write a function that takes a string as input and returns ...
- 2017/11/5 Leetcode 日记
2017/11/5 Leetcode 日记 476. Number Complement Given a positive integer, output its complement number. ...
- 2017/11/3 Leetcode 日记
2017/11/3 Leetcode 日记 654. Maximum Binary Tree Given an integer array with no duplicates. A maximum ...
- 2017.11.21 查询某个字段为null的记录
注意,不使用 = null, 而是 is null. select fd_username, fd_tenantid, fd_validity from t_user WHERE fd_validit ...
随机推荐
- CF864 E DP 输出路径
n个物品有Deadline,拿物品需要花费时间,问取得最大价值的方案. 本质是个01背包,先按时间排序,然后把花费的时间作为背包就行了. 主要就是找方案,倒过来找发生转移的就行了. 太菜了真的不会打C ...
- Uva5211/POJ1873 The Fortified Forest 凸包
LINK 题意:给出点集,每个点有个价值v和长度l,问把其中几个点取掉,用这几个点的长度能把剩下的点围住,要求剩下的点价值和最大,拿掉的点最少且剩余长度最长. 思路:1999WF中的水题.考虑到其点的 ...
- 一般处理程序、ASP.NET核心知识(5)
初窥 1.新建一个一般处理程序 新建一个一般处理程序 2.看看里头的代码 public class MyHandler : IHttpHandler { public void ProcessRequ ...
- XML & JSON---iOS-Apple苹果官方文档翻译
技术博客http://www.cnblogs.com/ChenYilong/ 新浪微博http://weibo.com/luohanchenyilong //转载请注明出处--本文永久链接 ...
- RabbitMQ使用简记
RabbitMQ是什么 MQ全称为Message Queue, 即消息队列.MQ是一种应用程序对应用程序的通信方法.应用程序通过读写出入队列的消息(针对应用程序的数据)来通信,而无需专用连接来链接它们 ...
- [SCOI2010]生成字符串 题解(卡特兰数的扩展)
[SCOI2010]生成字符串 Description lxhgww最近接到了一个生成字符串的任务,任务需要他把n个1和m个0组成字符串,但是任务还要求在组成的字符串中,在任意的前k个字符中,1的个数 ...
- 【leetcode 简单】第四题 罗马数字转整数
罗马数字包含以下七种字符:I, V, X, L,C,D 和 M. 字符 数值 I 1 V 5 X 10 L 50 C 100 D 500 M 1000 例如, 罗马数字 2 写做 II ,即为两个并列 ...
- 五. Jmeter--HTTP Cookie Manager
1. 添加HTTP Cookie Manager 2.添加登录login http,request info 和 HTTP Header Manager 中的信息是从fiddler中拿的, 至于hea ...
- 阿里分布式开源框架DUBBO 入门+ 进阶+ 项目实战视频教程
史诗级Java/JavaWeb学习资源免费分享 欢迎关注我的微信公众号:"Java面试通关手册"(坚持原创,分享各种Java学习资源,面试题,优质文章,以及企业级Java实战项目回 ...
- Linux下的压缩解压缩
Linux下最常用的打包程序就是tar了,使用tar程序打出来的包我们常称为tar包,tar包文件的命令通常都是以.tar结尾的.生成tar包后,就可以用其它的程序来进 行压缩了,所以首先就来讲讲t ...