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有人可以解释一下下面冒泡排序中第二个 for 循环的确切目的吗？我知道第一个循环正在查看数组的第“i”个整数，但第二个 for 循环到底在看什么？
请原谅我对这个话题的无知。我编码不到一周，对这个主题有些困惑。
void sort(int array[], int size) {
for(int i = 0, x = size - 1; i < x; i++) {
for(int j = 0; j < x - 1; j++) {
if(array[j] > array[j + 1]) {
int tmp = array[j];
array[j] = array[j + 1];
array[j + 1] = tmp;
}
}
}
}
最佳答案
考虑到您想实现冒泡排序，我猜您的第一个循环也是错误的，因为第一个循环告诉您对列表进行排序所需的遍数。在冒泡排序的情况下，它等于 Total Number of elements - 1 对 n 个元素的列表进行排序需要遍数 (n - 1)，因此 i 的值 如果我没记错的话，我个人认为必须从 1 开始。此外，就您将变量声明为需要时而言，您提供的代码片段不像 C 语言编码风格。
第二个循环基本上是为了减少比较(元素数量 - 通过 - 1)，在每次迭代之后，因为在每次通过时，我们将最大的元素放在右侧(逻辑上未排序的列表)。因此，由于该元素处于正确的位置，所以我们不必将其与其他元素进行比较。
4 3 2 1 Original List
3 2 1 4 Pass 1
-
Now since this above 4 is in it's rightful place
we don't need to compare it with other elements.
Hence we will start from the zeroth element and
compare two adjacent values, till 1 (for Pass 2)
Here comparison will take place between 3 and 2,
and since 3 is greater than 2, hence swapping
between 3 and 2, takes place. Now 3 is comapred
with 1, again since greater value is on left
side, so swapping will occur. Now 3 is not compared
with 4, since both these values are in their
rightful place.
2 1 3 4 Pass 2
-
Now since this above 3 is in it's rightful place
we don't need to compare it with other elements.
Hence we will start from the zeroth element and
compare two adjacent values, till 1 (for Pass 3)
Here only one comparison will occur, between
2 and 1. After swapping 2 will come to it's rightful
position. So no further comparison is needed.
1 2 3 4 Pass 3
Here the list is sorted, so no more comparisons, after Pass 3.
void bubbleSort(int *ptr, int size)
{
int pass = 1, i = 0, temp = 0;
for (pass = 1; pass < size - 1; pass++)
{
for (i = 0; i <= size - pass - 1; i++)
{
if (*(ptr + i) > *(ptr + i + 1))
{
temp = *(ptr + i);
*(ptr + i) = *(ptr + i + 1);
*(ptr + i + 1) = temp;
}
}
printf("Pass : %d\n", pass);
for (temp = 0; temp < size; temp++)
printf("%d\t", *(ptr + temp));
puts("");
}
}
关于c - 在冒泡排序中使用第二个for循环，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/15319520/