public class SelectionSort {
public static void selectionSort(int []arr) {
if (arr == null || arr.length < 2){
return;
}
for (int i = 0; i < arr.length - 1; i++){
int minIndex = i;
for (int j = i + 1; j < arr.length; j++){
minIndex = arr[minIndex] < arr[j] ? minIndex : j;
}
swap(arr, i, minIndex);
}
}
public static void swap(int[] arr, int i, int j){
arr[i] = arr[i] ^ arr[j];
arr[j] = arr[i] ^ arr[j];
arr[i] = arr[i] ^ arr[j];
}
}
问题：这一步搞错了
minIndex = arr[minIndex] < arr[j] ? minIndex : j;
public class InsertionSort {
public static void insertionSort(int []arr) {
if (arr == null || arr.length < 2){
return;
}
for(int i=0; i < arr.length; i++) {
for(int j = i-1; j >= 0 && arr[j] > arr[j+1]; j—){
swap(arr,j, j+1);
}
}
}
public static void swap(int[] arr, int i, int j){
arr[i] = arr[i] ^ arr[j];
arr[j] = arr[i] ^ arr[j];
arr[i] = arr[i] ^ arr[j];
}
}
public class BubbleSort {
public static void bubbleSort(int []arr) {
if (arr == null || arr.length < 2) {
return;
}
for (int e = arr.length - 1; e > 0; e- -) {
boolean flag = true;
for (int i= 0; i< e; i++) {
if (arr[i] > arr[i+1]) {
swap(arr, i, j);
flag = false;
}
}
if (flag) {
break;
}
}
}
public static void swap(int[] arr, int i, int j){
arr[i] = arr[i] ^ arr[j];
arr[j] = arr[i] ^ arr[j];
arr[i] = arr[i] ^ arr[j];
}
}
二分查找某个数是否存在
public class BSExist {
public static boolean exist (int []sortedArr, int num) {
if (sortedArr ==null || sortedArr.length == 0) {
return;
}
int L = 0;
int R = sortedArr.length - 1;
int min = 0;
while(L < R) {
mid = L + ((R - L) >> 1);
if (sortedArr[mid] == num) {
return true;
}
else if (sortedArr[mid] > num) {
R = mid - 1;
}
else{
L = mid + 1;
}
}
return sortedArr[L] == num;
}
}
问题：这一步差点忘了
return sortedArr[L] == num;
在arr上，找满足大于等于value的最左位置
public class BsNearLeft {
public static int nearestIndex(int []arr, int value) {
int L = 0;
int R = arr.length - 1;
int index = -1;
while (L < R) {
mid = L + ((R - L) >> 2);
if (arr[mid] >= value) {
index = value;
R = mid - 1;
}
else{
L = mid + 1;
}
}
return index;
}
}
在arr上，找满足小于等于value的最右位置
public class BSNearRight {
public static int nearestIndex(int []arr, int value) {
int L = 0;
int R = arr.length - 1;
int index = -1;
while (L < R) {
int mid = L + ((R - L) >> 1);
if (arr[mid] <= value){
index = mid;
L = mid + 1;
} else{
R = mid - 1;
}
}
return index;
}
}
Arr无序，任意两个相邻位置不等，返回一个局部最小的位置
public class BSAwesome {
public static int getLessIndex(int []arr) {
if (arr == null || arr.length == 0) {
return -1;
}
if (arr.length == 1 || arr[0] <arr[1]) {
return 0;
}
if (arr[arr.length - 2] > arr[arr.length - 1]) {
return arr.length - 1;
}
int L = 1;
int R = arr.length - 2;
int mid = 0;
while(L < R) {
mid = L + ((R - L) >> 1);
if (arr[mid] > arr[mid - 1]) {
R = mid - 1;
}
if (arr[mid] < arr[mid + 1]) {
L = mid + 1;
}else {
return mid;
}
}
return L;
}
}
问题：1、思考不全面
if (arr == null || arr.length == 0) {
return -1;
}
if (arr.length == 1 || arr[0] <arr[1]) {
return 0;
}
2、思考不全面
else {
return mid;
}
arr中，只有一种数，出现了奇数次，找出打印
arr中，只有两种数，出现了奇数次，找出打印
提取出一个数最右侧的1
public class EvenTimesOddTimes {
// arr中，只有一种数，出现了奇数次，找出打印
public static void printOddTimesNum1(int [arr]) {
int err = 0;
for (int i = 0; i < arr.length; i++) {
err ^= arr[i];
}
System.out.println(err);
}
// arr中，只有两种数，出现了奇数次，找出打印
public static void printOddTimesNum2(int []arr) {
int err = 0;
for (int i = 0; i < arr.length; i++) {
err ^= arr[i];
}
// 提取出一个数最右侧的1 N： 对N取反加1然后和N &运算
int rightOne = err & (～eor + 1);
int err2 = 0;
for (int i = 0; i < arr.length; i++) {
if (arr[i] & rightOne != 0) {
err2 ^= arr[i];
}
}
System.out.println(err2 + “” + err2 ^ err);
}
}
public class ReverseList {
public static class Node {
public int value;
public Node next;
public Node(int value) {
this.value = value;
}
}
public static void DoubleNode {
public int value;
public DoubleNode last;
public DoubleNode next;
public DoubleNode(int value){
this.value = value;
}
}
public static Node reverseLinkedList(Node head) {
Node pre = null;
Node next = null;
while(head != null) {
next = head.next;
head.next = pre;
pre = head;
head = next;
}
return pre;
}
public static DoubleNode reverseDoubleList(DoubleNode head) {
DoubleNode pre = null;
DoubleNode next = null;
while(head != null){
next = head.next;
head.next = pre;
head.last = next;
pre = head;
head = next;
}
return pre;
}
}
public class DeleteGivenValue{
public static class Node {
public int value;
public Node next;
public Node(int value) {
this.value = value;
}
}
public static Node removeValue(Node head, int num) {
while(head != null) {
if (head.value == num){
head = head.next;
}
}
pre = head;
cur = head;
while (cur != null) {
if (cur.value == num) {
pre.next = cur.next;
}else {
pre = cur;
}
cur = cur.next;
}
}
}
问题：不管相等不相等，cur都要来到下一个位置
while (cur != null) {
if (cur.value == num) {
pre.next = cur.next;
}else {
pre = cur;
}
cur = cur.next;
}
public class RingArray {
public static class myQueue {
private int[] arr;
private int pushi;
private int polli;
private int size;
private final int limit;
public myQueue(int limit) {
arr = new int[limit];
pushi = 0;
polli = 0;
this.limit = limit;
}
public void push(int value) {
if ( size == limit) {
throw new RuntimeException(“is MAX”);
}
size ++;
arr[pushi] = value;
pushi = nextIndex(pushi);
}
public int pop() {
if ( size == 0) {
throw new RuntimeException(“is null”);
}
size —;
int ans = arr[polli];
polli = nextIndex(polli);
}
public int nexIndex(int i) {
return i < limit - 1 ? i + 1 : 0;
}
}
}
public class GetMinStack {
public class MyStack1 {
private Stack<Integer> stackData;
private Stack<Integer> stackMin;
public MyStack1() {
this.stackData = new Stack<Integer>();
this.stackMin = new Stack<Integer>();
}
public void push(int newNum) {
if (this.stackMin.isEmpty()){
this.stackMin.push(newNum);
}else if (this.getMin() >= newNum) {
this.stackMin.push(newNum);
}
this.stackData.push(newNum);
}
public int pop() {
if (this.stackData.isEmpty()) {
throw new RuntimeException(“”);
}
int value = this.stackData.pop();
if (value == this.getMin()) {
this.stackMin.pop();
}
return value;
}
public int getMin() {
if (this.stackMin.isEmpty()) {
throw new RuntimeException(“stack is empty”);
}
return this.stackMin.peek();
}
}
}
public class TwoStacksImplementQueue {
public static class TwoStacksQueue {
public Stack<Integer> stackPush;
public Stack<Integer> stackPop;
public TwoStacksQueue() {
stackPush = new Stack<Integer>();
stackPop = new Stack<Integer>();
}
private void pushToPop() {
if (stackPop.empty()) {
while (!stackPush.empty()) {
stackPop.push(stackPush.Pop());
}
}
}
public void add (int PushInt) {
stackPush.push(PushInt);
pushToPop();
}
public int poll (){
if (stackPop.empty() && stackPush.empty()) {
throw new RuntimeException(“null”);
}
pushToPop();
return stackPop.pop();
}
public int peek() {
if (stackPop.empty() && stackPush.empty()) {
throw new RuntimeException(“null”);
}
pushToPop();
return stackPop.peek();
}
}
}
offer poll peek
public class TwoQueueImplementStack {
public static class TwoQueueStack<T> {
public Queue<T> queue;
public Queue<T> help;
public TwoQueueStack(T) {
queue = new LinkedList<T>();
help = new LinkedList<>();
}
public void push(T value) {
queue.offer(value);
}
public T poll() {
while (queue.size() > 1) {
help.offer(queue.poll());
}
T ans = queue.poll();
Queue<T> tmp = queue;
queue = help;
help = tmp;
return ans;
}
public T peek() {
while (queue.size() > 1) {
help.offer(queue.poll());
}
T ans = queue.poll();
help.offer(ans);
Queue<T> tmp = queue;
queue = help;
help = tmp;
return ans;
}
public boolean isEmpty() {
return queue.isEmpty();
}
}
}
public class MergeSort {
public static void mergeSort(int[] arr) {
if (arr == null || arr.length < 2) {
return;
}
process(arr, 0, arr.length - 1);
}
public static void process(int[] arr, int L, int R) {
if (L = R) {
return;
}
int mid = L + ((R - L) >> 1);
process(arr, L, mid);
process(arr, mid, R);
merge(arr, L, mid, R);
}
public static void merge(int[] arr, L, mid, R) {
int[] help = new int[R-L+1];
int i = 0;
int p1 = L;
int p2 = mid + 1;
while (p1 <= mid && p2 <= R) {
help[i++] = arr[p1] <= arr[p2] ? arr[p1++] : arr[p2++];
}
while (p1 <= mid) {
help[i++] = arr[p1++];
}
while (p2 <= R) {
help[i++] = arr[p2++];
}
for (i =0; i < help.length; i ++) {
arr[L + i] = help[i];
}
}
public static void main(String[] args) {}
}
public class SmallSum {
public static int smallSum(int[] arr) {
if (arr == null || arr.length < 2) {
return 0;
}
process(arr, 0, arr.length - 1);
}
public static int process(int[] arr, int L, int R) {
if (L=R) {
return 0;
}
int mid = l + ((r - l) >> 1);
return
process(arr, l, mid)
+
process(arr, mid + 1, r)
+
merge(arr, l, mid, r);
}
public static int merge(int[] arr, L, mid, R) {
int[] help = new int[R-L+1];
int i = 0;
int p1 = L;
int p2 = mid + 1;
int res = 0;
while (p1 <= mid && p2 <= R) {
res += arr[p1] < arr[p2] ? arr[p1] * (R - p2 + 1) : 0;
help[i++] = arr[p1] < arr[p2] ? arr[p1++] : arr[p2++];
}
while (p1 <= mid) {
help[i++] = arr[p1++];
}
while (p2 <= R) {
help[i++] = arr[p2++];
}
for (i =0; i < help.length; i ++) {
arr[L + i] = help[i];
}
}
}
public class ReversePair {
public static int reversePair(int[] arr) {
if (arr == null || arr.length < 2) {
return 0;
}
return process(arr, 0, arr.length - 1);
}
public static int process(int[] arr, int L, int R) {
if (L=R) {
return 0;
}
int mid = l + ((r - l) >> 1);
return
process(arr, l, mid)
+
process(arr, mid + 1, r)
+
merge(arr, l, mid, r);
}
public static int merge(int[] arr, L, mid, R) {
int[] help = new int[R-L+1];
int i = 0;
int p1 = L;
int p2 = mid + 1;
int res = 0;
while (p1 <= mid && p2 <= R) {
res += arr[p1] > arr[p2] ? mid - p1 + 1 : 0;
help[i++] = arr[p1] >= arr[p2] ? arr[p1++] : arr[p2++];
}
while (p1 <= mid) {
help[i++] = arr[p1++];
}
while (p2 <= R) {
help[i++] = arr[p2++];
}
for (i =0; i < help.length; i ++) {
arr[L + i] = help[i];
}
}
}
public class QuickSort {
public static void quickSort(int[] arr) {
if (arr == null || arr.length < 2) {
return;
}
process3(arr, 0, arr.length - 1);
}
public static void process3(int[] arr, int L, int R) {
if (L >= R) {
return;
}
swap(arr, L + (int) (Math.random() * (R - L + 1)), R);
int[] equalArea = partition(arr, L, R);
process3(arr, L, equalArea[0] - 1);
process3(arr, equalArea[1] + 1, R);
}
public static int[] partition(int[] arr, int L, int R) {
if (L > R) { // L...R L>R
return new int[] { -1, -1 };
}
if (L == R) {
return new int[] { L, R };
}
int less = L - 1; // < 区 右边界
int more = R; // > 区 左边界
int index = L;
while(index < more) {
if (arr[index] == arr[R]) {
index++;
} else if (arr[index] < arr[R]) {
swap(arr, index++, ++less);
} else {
swap(arr, index, --more);
}
}
swap(arr, more, R);
return new int[] { less + 1, more };
}
}
public class HeapSort {
public static void heapSort(int[] arr) {
if (arr == null || arr.length < 2) {
return;
}
// O(N)
for (int i = arr.length - 1; i >= 0; i--) {
heapify(arr, i, arr.length);
}
int heapSize = arr.length;
swap(arr, 0, --heapSize);
// O(N*logN)
while (heapSize > 0) { // O(N)
heapify(arr, 0, heapSize); // O(logN)
swap(arr, 0, --heapSize); // O(1)
}
}
public static void heapInsert(int[] arr, int index) {
while (arr[index] > arr[(index - 1) / 2]) {
swap(arr, index, (index - 1) / 2);
index = (index - 1) / 2;
}
}
public static void heapify(int[] arr, int index, int heapSize) {
int left = index * 2 + 1;
while (left < heapSize) {
int largest = left + 1 < heapSize && arr[left + 1] > arr[left] ? left + 1 : left;
largest = arr[largest] > arr[index] ? largest : index;
if (largest == index) {
break;
}
swap(arr, largest, index);
index = largest;
left = index * 2 + 1;
}
}
public static void swap(int[] arr, int i, int j) {
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
}
}
public class SortArrayDistanceLessK {
public static void sortedArrDistanceLessK(int[] arr, int k) {
if (k == 0) {
return;
}
// 默认小根堆
PriorityQueue<Integer> heap = new PriorityQueue<>();
int index = 0;
// 0...K
for (; index <= Math.min(arr.length - 1, k); index++) {
heap.add(arr[index]);
}
int i = 0;
for (; index < arr.length; i++, index++) {
heap.add(arr[index]);
arr[i] = heap.poll();
}
while (!heap.isEmpty()) {
arr[i++] = heap.poll();
}
}
}
public class RecursiveTraversalBT {
public static class Node {
public int value;
public Node left;
public Node right;
public Node(int v) {
value = v;
}
}
public static void f(Node head) {
if (head == null) {
return;
}
// 1
f(head.left);
// 2
f(head.right);
// 3
}
}
public class UnRecursiveTraversalBT {
public static class Node {
public int value;
public Node left;
public Node right;
public Node(int v) {
}
// 先序遍历
public static void pre(Node head) {
if (head != null) {
Stack<Node> stack = new Stack<Node>();
stack.push(head);
if (!stack.isEmpty()) {
head = stack.pop();
System.out.print(head.value + " ");
if (head.right != null) {
stack.push(head.right);
}
if (head.left != null) {
stack.push(head.left);
}
}
}
}
// 中序遍历 左边界
public static void in(Node cur) {
if (cur != null) {
Stack<Node> stack = new Stack<Node>();
while (!stack.isEmpty() || cur != null) {
if (cur != null) {
stack.push(cur);
cur = cur.left;
}else{
cur = stack.pop();
System.out.print(cur.value + " ");
cur = cur.right;
}
}
}
}
// 后序遍历
public static void pos1(Node head) {
if (head != null) {
Stack<Node> s1 = new Stack<Node>();
Stack<Node> s2 = new Stack<Node>();
s1.push(head);
while(!s1.isEmpty()) {
head = s1.pop();
s2.push(head);
if (head.left != null) {
s1.push(head.left);
}
if (head.right != null) {
s1.push(head.right);
}
}
}
while (!s2.isEmpty()) {
System.out.print(s2.pop().value + " ");
}
}
//使用一个栈实现后序遍历
public static void pos2(Node h) {
if (h != null) {
Stack<Node> stack = new Stack<Node>();
stack.push(h);
Node c = null;
while (!stack.isEmpty()) {
c = stack.peek();
if (c.left != null && h != c.left && h != c.right) {
stack.push(c.left);
} else if (c.right != null && h != c.right) {
stack.push(c.right);
} else {
System.out.print(stack.pop().value + " ");
h = c;
}
}
}
System.out.println();
}
}
public class LevelTraversalBT {
public static class Node {
public int value;
public Node left;
public Node right;
public Node(int v) {
value = v;
}
}
public static void level(Node head) {
if (head == null) {
return;
}
Queue<Node> queue = new LinkedList<>();
queue.add(head);
while (!queue.isEmpty()) {
Node cur = queue.poll();
System.out.println(cur.value);
if (cur.left != null) {
queue.add(cur.left);
}
if (cur.right != null) {
queue.add(cur.right);
}
}
}
}
public class TreeMaxWidth {
public static class Node {
public int value;
public Node left;
public Node right;
public Node(int data) {
this.value = data;
}
}
public static int maxWidthUseMap(Node head) {
if (head == null) {
return 0;
}
Queue<Node> queue = new LinkedList<>();
queue.add(head);
// key 在 哪一层，value
HashMap<Node, Integer> levelMap = new HashMap<>();
levelMap.put(head, 1);
int curLevel = 1; // 当前你正在统计哪一层的宽度
int curLevelNodes = 0; // 当前层curLevel层，宽度目前是多少
int max = 0;
while (!queue.isEmpty()) {
Node cur = queue.poll();
int curNodeLevel = levelMap.get(cur);
if (cur.left != null) {
levelMap.put(cur.left, curNodeLevel + 1);
queue.add(cur.left);
}
if (cur.right != null) {
levelMap.put(cur.right, curNodeLevel + 1);
queue.add(cur.right);
}
if (curNodeLevel == curLevel) {
curLevelNodes++;
} else {
max = Math.max(max, curLevelNodes);
curLevel++;
curLevelNodes = 1;
}
}
max = Math.max(max, curLevelNodes);
return max;
}
public static int maxWidthNoMap(Node head) {
if (head == null) {
return 0;
}
Queue<Node> queue = new LinkedList<>();
queue.add(head);
Node curEnd = head; // 当前层，最右节点是谁
Node nextEnd = null; // 下一层，最右节点是谁
int max = 0;
int curLevelNodes = 0; // 当前层的节点数
while (!queue.isEmpty()) {
Node cur = queue.poll();
if (cur.left != null) {
queue.add(cur.left);
nextEnd = cur.left;
}
if (cur.right != null) {
queue.add(cur.right);
nextEnd = cur.right;
}
curLevelNodes++;
if (cur == curEnd) {
max = Math.max(max, curLevelNodes);
curLevelNodes = 0;
curEnd = nextEnd;
}
}
return max;
}
}
public class SuccessorNode {
public static class Node {
public int value;
public Node left;
public Node right;
public Node parent;
public Node(int data) {
this.value = data;
}
}
public static Node getSuccessorNode(Node node) {
if (node == null) {
return node;
}
if (node.right != null) {
return getLeftMost(node.right);
} else { // 无右子树
Node parent = node.parent;
while (parent != null && parent.right == node) { // 当前节点是其父亲节点右孩子
node = parent;
parent = node.parent;
}
return parent;
}
}
public static Node getLeftMost(Node node) {
if (node == null) {
return node;
}
while (node.left != null) {
node = node.left;
}
return node;
}
}
一日之计在于晨，来一发！补充功能：效果图demo源码 tab选项卡+CoordinatorLayout收缩布局+复杂Recyclerview列表学习的步伐（六） Kotlin 学习总结：类的特性 学习的步伐（五） Kotlin 基础语法学习总结：语法 学习的步伐（四） Kotlin 基础语法学习总结：操作符 学习的步伐（三）Kotlin TabLayout+Viewpager+Frag
原理：目标物体的轮廓实质是指一系列像素点构成，这些点构成了一个有序的点集。我们可以通过findContours函数将二值图像的边缘像素点分成多个轮廓，从而逐个提取目标外部轮廓，内部轮廓有待研究。Python：import cv2 as cvimport numpy as npif __name__=="__main__": img=cv.imread("D:/testimage...
扩展串口一、为什么要扩展串口？一块单片机的串口是有限的，一般2~4个。当我们做一个项目时需要连接多个外设时跟单片机通讯时，且通讯都是以串口形式。那么我们只能去扩展串口来满足我们的应用需求。二、解决方法1、选择拥有更多UART芯片。2、外部添加接口转换芯片。SP2538芯片，它可轻松的将任意单片机（如89C51）或DSP等现有的RS232串口扩展成5个全新的全新的全双工RS232串行口（所有串口可独立接收数据），具体使用可查询芯片手册。3、选择RS485的外设代替RS232外.
转载请标明出处： http://blog.csdn.net/forezp/article/details/51873137 本文出自方志朋的博客最近在研究android 开发的新控件，包括drawer layout ,NavigationView,CoordinatorLayout,AppBarLayout,T...
实验中需要用到区域联通的算法，就是类似于matlab中bwlabel的函数。网上找了找c++源码未果，bwlabel-python版用python描述了matlab中的实现方法，但是最后对标签的处理部分并未看明白，故自己用c++实现了一个。先直接看bwlabel函数代码：cv::Mat bwlabel(const cv::Mat in, int * num, const int mode){
似乎是noip2017d2t3的一个部分分。用splay的话当然非常裸，但说不定会被卡常。可以发现序列中数的（环上）相对位置是不变的，考虑造一棵权值线段树维护权值区间内还有多少个数留在序列中，每次在线段树上二分即可。#include<iostream> #include<cstdio>#include<cmath>#include<...
给大家几个建立用例模型中常出现的问题和应对遵循的原则： 一．如何发现用例 经过以上的讲解，相信大家对建立用例模型有了一个整体的概念，然后开始着手练习绘制用例模型。这时候，一个非常严峻的问题出现了：如何发现用例。大师曾经给出了答案，大致意思就是：首先选择系统边界，然后确定主要参与者，定义满足用户目标的用例，为其命名。然而，我在实践中证明，这套方法过于理论，并不实用。也许，我们...
按照官方手册http://www.kancloud.cn/manual/thinkphp5/142357 折腾了很久还是无法实现，可能还是我理解的不对，最后使用了如下方式：1. 项目表DROP TABLE IF EXISTS `darling_project`;CREATE TABLE `darling_project` ( `project_id` int(32) N
目录文章目录目录AWS Direct ConnectAWS Direct Connect 的优势稳定的网络性能降低带宽成本保护传输中的数据AWS Direct Connect 的网络架构Direct Connect vs VPC IPSec VPN ConnectionsAWS Direct ConnectAWS Direct Connect 是一种用于替代 Internet 来连接到 AWS Cloud 的网络服务，由 AWS 或 AWS Direct Connect 的 APN 合作伙伴提供服务。
京东CTO张晨卸任。今天下午，京东集团宣布，原京东集团首席技术官(CTO)张晨由于家庭原因需长期在海外生活，他将卸任京东集团首席技术官(CTO)，同时从2019年6月30日起担任集团顾问，京东将在未来宣布相关继任计划。张晨表示，“我很骄傲过去能够成为京东集团决策层的一员，也很荣幸能在京东转型成长的过程中担任首席技术官。因为个人和家庭原因的规划考虑，很遗憾我无法继续全职在中国工作，但我还是很...
目录一、中断相关的基础概念二、CC2530的中断系统三、CC2530的中断处理函数编写方法四、CC2530的外部中断五、实训案例：外部中断输入控制LED灯一、中断相关的基础概念 内核与外设之间的主要交互方式有两种：轮询和中断。轮询的方式貌似公平，但实际工作效率很低，且不能及时响应紧急事件；中断系统使得内核具备了应对突发事件的能力。 在执行CPU当前程序时，由于系统中出现了某种急需处理的情况，CPU暂停正在执行的程序，转而去执行另外一段特殊程序来处...
水的一天......看了状压和一部分概率疲惫期