Bubble Sort，bubblesort

8 numbers. Sort as ascend.

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1st loop, compare 7 times (for 8 numbers), and found the largest 8.

2nd loop, compare 6 times (for 7 numbers), and found the largest 7.

. . .

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1, 7, 8

2, 6, 7

3, 5, 6

4, 4, 5

5, 3, 4

6, 2, 3

7, 1, 2

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In conclusion:?For sorting 8 numbers, we need an outer loop of 7 times, each time for finding a largest number; and an inner loop from comparing 7 times to comparing 1 time (as in the center column).

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Implementation in PHP:?

<span> 1</span> <?<span>php
</span><span> 2</span> <span>/*</span><span> bubble sort: 
</span><span> 3</span> <span>    1. operate directly on the input array (&), not on a copy
</span><span> 4</span> <span>    2. sort as ascend
</span><span> 5</span> 
<span> 6</span> <span>    a is array
</span><span> 7</span> <span>    m is length of a
</span><span> 8</span> <span>    n is times of outer loop, n-i is times of comparing for each outer loop
</span><span> 9</span> <span>    i/j is for-loop counter
</span><span>10</span> <span>    w is for value swap
</span><span>11</span> <span>*/</span>
<span>12</span> <span>function</span> sortBubble(&<span>$a</span><span>){
</span><span>13</span>     <span>$m</span> = <span>count</span>(<span>$a</span><span>);
</span><span>14</span>     <span>$n</span> = <span>$m</span> - 1<span>;
</span><span>15</span>     <span>for</span>(<span>$i</span>=0; <span>$i</span><<span>$n</span>; <span>$i</span>++<span>){
</span><span>16</span>         <span>for</span>(<span>$j</span>=0; <span>$j</span><<span>$n</span>-<span>$i</span>; <span>$j</span>++<span>){
</span><span>17</span>             <span>if</span>(<span>$a</span>[<span>$j</span>] > <span>$a</span>[<span>$j</span>+1<span>]){
</span><span>18</span>                 <span>$w</span> = <span>$a</span>[<span>$j</span><span>];
</span><span>19</span>                 <span>$a</span>[<span>$j</span>] = <span>$a</span>[<span>$j</span>+1<span>];
</span><span>20</span>                 <span>$a</span>[<span>$j</span>+1] = <span>$w</span><span>;
</span><span>21</span> <span>            }
</span><span>22</span>             <span>else</span><span>{
</span><span>23</span>                 <span>//</span><span> do nothing</span>
<span>24</span> <span>            }
</span><span>25</span> <span>        }
</span><span>26</span>         <span>//</span><span> see the results after each outer loop
</span><span>27</span> <span>        // echo implode(', ', $a).'<br />';</span>
<span>28</span> <span>    }
</span><span>29</span> <span>}
</span><span>30</span> 
<span>31</span> <span>$arr</span> = <span>array</span>(9, 5, 2, 7, 3<span>);
</span><span>32</span> sortBubble(<span>$arr</span><span>);
</span><span>33</span> <span>echo</span> <span>implode</span>(', ', <span>$arr</span><span>);
</span><span>34</span> 
<span>35</span> <span>//</span><span> 2, 3, 5, 7, 9</span>
<span>37</span> ?>

?

冒泡排序c

冒泡排序詳細(xì)注釋：
/* 用冒泡排序法對(duì)一維整型數(shù)組中的十個(gè)數(shù)升序排序 */
#include
#include

int main()
{
int i,j,t,a[10];
printf("Please input 10 integers:\n");
for(i=0;iscanf("%d",&a[i]);
for(i=0;ifor(j=0;jif(a[j]>a[j+1])
{t=a[j];/* 交換a[i]和a[j] */
a[j]=a[j+1];
a[j+1]=t;
}
printf("The sequence after sort is:\n");
for(i=0;iprintf("%-5d",a[i]);
printf("\n");
system("pause");
return 0;
}
其中i=0時(shí)：
j從0開始a[0],a[1]比較大小，把其中的較大者給a[1],然后j++,a[1]和a[2]再比較，再把兩者中的
較大者給a[2],這樣a[0],a[1],a[2]中的最大者已經(jīng)交換到a[2]中，這個(gè)過程繼續(xù)，直到j(luò)=10-i-1=9這樣
a[9]中的為10個(gè)數(shù)中的最大數(shù)。
然后i=1時(shí):
由于最大數(shù)已找到并放到a[9]中，所以這一次循環(huán)j最大只需到10-i-1=8，即a[8]即可，再次從j=0開始a[j]和a[j+1]兩兩比較交換，最后次大數(shù)放到a[8]中
然后i++,繼續(xù)...
當(dāng)i=9時(shí)已經(jīng)過9次兩兩比較完成所有排序，i對(duì)于n個(gè)數(shù)，只需要進(jìn)行n-1次外循環(huán)的兩兩比較就完成排序。
至于按降序排列只需將if(a[j]>a[j+1])改為if(a[j]
-------------------------------------------------------------------
/* 用改進(jìn)型冒泡排序法對(duì)一維整型數(shù)組中的十個(gè)數(shù)升序排序 */
#include
#include
int main()
{int i,j,t,a[10],flag;
printf("Please input 10 integers:\n");
for(i=0;iscanf("%d",&a[i]);
for(i=0;i{ flag=0;
for(j=0;jif(a[j]>a[j+1])
{ t=a[j]; /* 交換a[i]和a[j] */
a[j]=a[j+1];
a[j+1]=t;
flag=1;
}
if(flag==0)break;
}
printf("The sequence after sort is:\n&......余下全文>>
?

冒泡排序c

冒泡排序詳細(xì)注釋：
/* 用冒泡排序法對(duì)一維整型數(shù)組中的十個(gè)數(shù)升序排序 */
#include
#include

int main()
{
int i,j,t,a[10];
printf("Please input 10 integers:\n");
for(i=0;iscanf("%d",&a[i]);
for(i=0;ifor(j=0;jif(a[j]>a[j+1])
{t=a[j];/* 交換a[i]和a[j] */
a[j]=a[j+1];
a[j+1]=t;
}
printf("The sequence after sort is:\n");
for(i=0;iprintf("%-5d",a[i]);
printf("\n");
system("pause");
return 0;
}
其中i=0時(shí)：
j從0開始a[0],a[1]比較大小，把其中的較大者給a[1],然后j++,a[1]和a[2]再比較，再把兩者中的
較大者給a[2],這樣a[0],a[1],a[2]中的最大者已經(jīng)交換到a[2]中，這個(gè)過程繼續(xù)，直到j(luò)=10-i-1=9這樣
a[9]中的為10個(gè)數(shù)中的最大數(shù)。
然后i=1時(shí):
由于最大數(shù)已找到并放到a[9]中，所以這一次循環(huán)j最大只需到10-i-1=8，即a[8]即可，再次從j=0開始a[j]和a[j+1]兩兩比較交換，最后次大數(shù)放到a[8]中
然后i++,繼續(xù)...
當(dāng)i=9時(shí)已經(jīng)過9次兩兩比較完成所有排序，i對(duì)于n個(gè)數(shù)，只需要進(jìn)行n-1次外循環(huán)的兩兩比較就完成排序。
至于按降序排列只需將if(a[j]>a[j+1])改為if(a[j]
-------------------------------------------------------------------
/* 用改進(jìn)型冒泡排序法對(duì)一維整型數(shù)組中的十個(gè)數(shù)升序排序 */
#include
#include
int main()
{int i,j,t,a[10],flag;
printf("Please input 10 integers:\n");
for(i=0;iscanf("%d",&a[i]);
for(i=0;i{ flag=0;
for(j=0;jif(a[j]>a[j+1])
{ t=a[j]; /* 交換a[i]和a[j] */
a[j]=a[j+1];
a[j+1]=t;
flag=1;
}
if(flag==0)break;
}
printf("The sequence after sort is:\n&......余下全文>>
?