鉴于代码:
for (int i = 0; i < n; ++i)
{
A(i) ;
B(i) ;
C(i) ;
}
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和优化版本:
for (int i = 0; i < (n - 2); i+=3)
{
A(i)
A(i+1)
A(i+2)
B(i)
B(i+1)
B(i+2)
C(i)
C(i+1)
C(i+2)
}
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我不清楚:哪个更好?我看不到任何使用其他版本的速度更快的东西.我在这里错过了什么吗?
我只看到每条指令都依赖于前面的指令,这意味着我需要等待前一条指令完成才能启动后面的指令...
谢谢
如何计算C字符串中出现的次数/?
我可以做这个:
int countSlash(char str[])
{
int count = 0, k = 0;
while (str[k] != '\0')
{
if (str[k] == '/')
count++;
k++;
}
return count;
}
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但这不是一种优雅的方式; 有关如何改进它的任何建议?
我有一个histShape.m带函数的文件histShape,还有一些其他函数.
代码的一般视图是:
%
function [outputImage] = histShape(srcimg, destimg)
PIXELS = 255 + 1;
....
....
end
%
function [outputImage] = normalizeAndAccumulate(inputImage)
PIXELS = 255 + 1;
....
....
end
%
function [pixels] = getNormalizedHistogram(histogram , inputImage)
PIXELS = 255 + 1;
....
....
end
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我可以使用,global x y z; 但我正在寻找一种不同的方式.
我想将变量声明PIXELS为全局变量,我该怎么做?
问候
假设这srcHoughMatrix是一个三维矩阵:
代替
if (currentRadius >= MINIMUM_ALLOWED_RADIUS )
% we're using only radiuses that are 6 or above
currentHough = srcHoughMatrix(index,jindex,currentRadius);
srcHoughMatrix(index,jindex,currentRadius) = currentHough + 1;
end
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1如果条件为真,我如何添加到每个单元格,而不使用临时变量或不使用临时变量
srcHoughMatrix(index,jindex,currentRadius) = srcHoughMatrix(index,jindex,currentRadius) + 1;
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谢谢
鉴于以下内容:
public class NavigationCanvas extends Canvas implements MouseListener,MouseMotionListener,KeyListener {
public void paint(Graphics g)
{
// some code
// more
// ...
g.setColor(Color.black);
// drawing each Line
for (int i=0; i<length; i++)
{
Line2D currLine = m_lines.get(i);
g.drawLine((int)currLine.getX1(),(int)currLine.getY1(),
(int)currLine.getX2(),(int)currLine.getY2());
g.drawLine((int)currLine.getX1()+1,(int)currLine.getY1()+1
,(int)currLine.getX2()+1,(int)currLine.getY2()+1);
g.drawLine((int)currLine.getX1()+2,(int)currLine.getY1()+2
,(int)currLine.getX2()+2,(int)currLine.getY2()+2);
}
}
...
}
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当我绘制线条时,currLine我得到这个:

正如你所看到的,我打了3个电话drawline(),让它更大胆,但它仍然不如我想要的那样.
我怎样画一条粗线?
我已经看了这个问题,但我仍然没有看到后缀树和Trie之间的区别.
两者都具有给定字符串的所有子字符串,因此它们彼此之间有何不同?
我使用“react-dropzone”来删除文件:
import React from "react";
import { useDropzone } from "react-dropzone";
const DropzoneUpload = ({ onDrop, accept }) => {
// Initializing useDropzone hooks with options
const { getRootProps, getInputProps, isDragActive } = useDropzone({
onDrop,
accept
});
return (
<div {...getRootProps()}>
<input className="dropzone-input" {...getInputProps()} />
<div className="text-center">
{isDragActive ? (
<p className="dropzone-content">Release to drop the files here</p>
) : (
<p className="dropzone-content">
Drag 'n' drop some files here, or click to select files
</p>
)}
</div>
</div>
);
};
export …Run Code Online (Sandbox Code Playgroud) 我正在学习使用 Material UI Grid,我想在列的右侧添加一个空白列(例如从第一个元素右侧填充),而不使用填充。
考虑代码:
import React from 'react';
import Grid from "@material-ui/core/Grid";
import Paper from "@material-ui/core/Paper";
import { makeStyles } from '@material-ui/core/styles';
const ExampleGridComponent = () => {
const useStyles = makeStyles(theme => ({
root: {
flexGrow: 1,
},
paper: {
padding: theme.spacing(2),
textAlign: 'center',
color: theme.palette.text.secondary,
},
}));
const classes = useStyles();
return (
<Grid container spacing={3}>
<Grid item xs={12}>
<Paper className={classes.paper}>xs=12</Paper>
</Grid>
<Grid item xs={12} sm={6}>
<Paper className={classes.paper}>xs=12 sm=6</Paper>
</Grid>
<Grid item xs={12} sm={6}>
<Paper className={classes.paper}>xs=12 sm=6</Paper> …Run Code Online (Sandbox Code Playgroud) 我在以下位置创建了一个 Switch Hyper-V Manager :
Virtual Switch Manager (on the right) => New Virtual Network Switch => External => MinikubeSwitch
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进而 :
在 CLI 中(在 Windows 10 中的管理员下)点击该行后:
minikube start --driver=hyperv --hyperv-virtual-switch=MinikubeSwitch
I got :
PS C:\bin> minikube start --driver=hyperv --hyperv-virtual-switch=MinikubeSwitch
* minikube v1.9.2 on Microsoft Windows 10 Pro 10.0.18363 Build 18363
* Using the hyperv driver based on user configuration
* Starting control plane node m01 in cluster minikube
* Creating hyperv VM (CPUs=2, Memory=2200MB, Disk=20000MB) ... ! StartHost …Run Code Online (Sandbox Code Playgroud) c ×2
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awt ×1
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