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
- ② 设计模式的艺术-08.桥接(Bridge)模式
为什么需要桥接(Bridge)模式 商城系统中常见的商品分类,以电脑为类,如何良好的处理商品分类销售的问题? 采用多层继承结构: 多层继承结构代码示例 Computer.java package co ...
- VS调试程序快捷键和系统快捷键
调试程序快捷键 编译程序:F7 运行程序:ctrl + F5 打断点:F9 运行到断点位置:F5 单步执行:F10 单步进入函数:F11 结束调试:shift+F5 注释代码:ctrl+k,ctrl+ ...
- .NET中的异常和异常处理
.NET中的异常(Exception) .net中的中异常的父类是Exception,大多数异常一般继承自Exception. 可以通过编写一个继承自Exception的类的方式,自定义异常类! 异常 ...
- Tomcat面试题目
1.tomcat给你你怎样去调优? 1. JVM参数调优:-Xms<size> 表示JVM初始化堆的大小,-Xmx<size>表示JVM堆的最大值.这两个值的大小一般根据需要进 ...
- 信息收集之zoomeye
一.浏览器上使用api接口 1.https://api.zoomeye.org/user/login post传参:{"username" : "username&quo ...
- # 2018高考&自主招生 - 游记
准备了一整个学期的高考和自主招生终于结束了....从育英出来, 以一个失败者的身份来写游记, 权当为明年的决战提供经验与总结. Day -1, June 5th 下午同学收拾考场, 自己在那里看书.. ...
- div遮罩实现禁用鼠标(click、hover等)事件
这两天在帮老师做网页,今天想实现在一块区域内禁止鼠标的各种事件,本来是想在框架模板的js文件里去修改,但是改代码的时候有点凌乱...感觉应该自己把问题想复杂了. 所以想了想要是能实现在一个区域内(如: ...
- 分布式系统的负载均衡以及ngnix负载均衡的五种策略
一般而言,有以下几种常见的负载均衡策略: 一.轮询. 特点:给每个请求标记一个序号,然后将请求依次派发到服务器节点中,适用于集群中各个节点提供服务能力等同且无状态的场景. 缺点:该策略将节点视为等同, ...
- Netty并发优化之ExecutionHandler
上文<Netty框架入门>说到:如果业务处理handler耗时长,将严重影响可支持的并发数. 针对这一问题,经过学习,发现了可以使用ExecutionHandler来优化. 先来回顾一下没 ...
- IT行业经典面试技巧及方法思路。
问题1:为什么从上家公司离职?”能说说原因吗? 首先,作为一个从事招聘的HR,并不认为追问面试者为什么从上一家公司离职是个明智的做法起码不应该在面试一开始就抛出这个问题,一个较为明显的原因是因为这会引 ...