Question

#include <stdio.h>

int main(void) {

    int a[5]={1,2,0,5,4},i,j,c;
    for(i=0;i<5;i++)    
    {
        for(j=0;j<5-i;j++){
            if(a[j]>a[j+1])
            {
                c=a[j];
                a[j]=a[j+1];
                a[j+1]=c;
            }
        }
    }
    for(i=0;i<5;i++)
    {
    printf("%d",a[i]);
    }
    return 0;
}

ideone说“超出时间限制：5个记忆：2048信号：24”

但它在turbo编译器上工作正常

Answer 1

for(j=0;j<5-i;j++){
            if(a[j]>a[j+1])

a[j+1]是数组越界访问，这将导致未定义的行为

Answer 2

尝试使用冒泡排序。在任何编译器中都不会妨碍机器循环。

for (i = 0; i < count -1; i++)
      for (j = 0; j < ((count - 1) - i); j++) {
          if (memptr[j]>memptr[j+1]) {
              temp = memptr[j];
              memptr[j] = memptr[j + 1];
              memptr[j + 1] = temp;
          }
      }

Answer 3

this.stop();
var mainStage = this;

var movieWidth = 640;//Choose Scaled Width of Video
var movieHeight = 360;//Choose Scaled Height of Video
var autoplayVideo = true;//Make it Autoplay

var vidya = document.createElement('video');//Creates the Video Element that is referenced later

var canvasEle = document.getElementById('canvas');//Identify the Canvas element
ctx = canvasEle.getContext('2d');//get canvas's context

canvasEle.parentNode.insertBefore(vidya, canvasEle);//insert video element before canvas element

vidya.setAttribute("src", "testing.mp4");//Place URL of the Video here (Local in this Example)
vidya.setAttribute("type","video/mp4");//What type of Video it is, Should be MP4
vidya.setAttribute("width", movieWidth);//scales the video to the width you had set above
vidya.setAttribute("controls","");//Turns on the default generic video controls
vidya.setAttribute("id","VIDEO");//gives the element an id for reference(Not Used yet)
if (autoplayVideo == true){ vidya.setAttribute("autoplay","");};

createjs.Ticker.addEventListener("tick", handleTick);

function handleTick(event){
    ctx.drawImage(vidya, 30, 70, movieWidth, movieHeight);
    console.log(ctx + " "+v);//here is where the uncaught reference is the "v"
}

因此，对于以下索引，您将获得以下内容：

int a[5]={1,2,0,5,4}

现在在你的外循环index ->| 0 | 1 | 2 | 3 | 4 | content ->| 1 | 2 | 0 | 5 | 4 |中，对于第一次迭代for(i=0;i<5;i++)，内部循环破坏条件将是，

i = 0

，或者

for(j=0;j<5-i;j++)

因此for(j=0;j<5-0;j++)的价值会增加j到0。

现在想想4时会发生什么a[j+1]！

您正尝试访问j = 4或a[4+1]，其中数组a[5]的{{1}}定义为index。

因此，在a，您将获得Undefined behavior。

尝试：

气泡排序超过了一个时间限制

3 个答案: