1. 冒泡排序法说明
- 冒泡法
- 属于交换排序
- 两两比较大小,交换位置。如同水泡咕嘟咕嘟往上冒
- 结果分为升序和降序排列
- 升序
- n个数从左至右,编号从0开始到n-1,索引0和1的值比较,如果索引0大,则交换两者位置,如果索引1大,则不交换。继续比较索引1和2的值,将大值放在右侧。直至n-2和n-1比较完,第一轮比较完成。第二轮从索引0比较到n-2,因为最右侧n-1位置上已经是最大值了。依次类推,每一轮都会减少最右侧的不参与比较,直至剩下最后2个数比较。
- 降序
2. 冒泡排序法举例
初始 | 1 9 8 5 6 7 4 3 2 | | | |
---|
第一趟 | 1 8 9 5 6 7 4 3 2 | 1 8 5 9 6 7 4 3 2 | 1 8 5 6 9 7 4 3 2 | 1 8 5 6 7 9 4 3 2 |
| 1 8 5 6 7 4 9 3 2 | 1 8 5 6 7 4 3 9 2 | 1 8 5 6 7 4 3 2 9 | |
第二趟 | 1 8 5 6 7 4 3 2 9 | 1 5 8 6 7 4 3 2 9 | 1 5 6 8 7 4 3 2 9 | 1 5 6 7 8 4 3 2 9 |
| 1 5 6 7 4 8 3 2 9 | 1 5 6 7 4 3 8 2 9 | 1 5 6 7 4 3 2 8 9 | |
第三趟 | 1 5 6 7 4 3 2 8 9 | 1 5 6 4 7 3 2 8 9 | 1 5 6 4 3 7 2 8 9 | 1 5 6 4 3 2 7 8 9 |
第四趟 | 1 5 6 4 3 2 7 8 9 | 1 5 4 6 3 2 7 8 9 | 1 5 4 3 6 2 7 8 9 | 1 5 4 3 2 6 7 8 9 |
第五趟 | 1 5 4 3 2 6 7 8 9 | 1 4 5 3 2 6 7 8 9 | 1 4 3 5 2 6 7 8 9 | 1 4 3 2 5 6 7 8 9 |
第六趟 | 1 4 3 2 5 6 7 8 9 | 1 3 4 2 5 6 7 8 9 | 1 3 2 4 5 6 7 8 9 | |
第七趟 | 1 3 2 4 5 6 7 8 9 | 1 2 3 4 5 6 7 8 9 | | |
第八趟 | 1 2 3 4 5 6 7 8 9 | | | |
3. 冒泡法代码实现(一)
m_list = [
[1, 9, 8, 5, 6, 7, 4, 3, 2],
[1, 2, 3, 4, 5, 6, 7, 8, 9],
[9, 8, 7, 6, 5, 4, 3, 2, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 2]
]
nums = m_list[0]
length = len(nums)
print(nums)
count_swap = 0
count_iter = 0
for i in range(length):
for j in range(length -i -1):
count_iter += 1
if nums[j] > nums[j + 1]:
nums[j], nums[j + 1] = nums[j + 1], nums[j]
count_swap +=1
print(nums, count_swap, count_iter)
4. 冒泡法代码实现(二)
m_list = [
[1, 9, 8, 5, 6, 7, 4, 3, 2],
[1, 2, 3, 4, 5, 6, 7, 8, 9],
[9, 8, 7, 6, 5, 4, 3, 2, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 2]
]
nums = m_list[0]
length = len(nums)
print(nums)
flag = False
count_swap = 0
count_iter = 0
for i in range(length):
for j in range(length -i -1):
count_iter += 1
if nums[j] > nums[j + 1]:
nums[j], nums[j + 1] = nums[j + 1], nums[j]
count_swap +=1
flag = True
if not flag:
break
print(nums, count_swap, count_iter)
m_list = [
[1, 9, 8, 5, 6, 7, 4, 3, 2],
[1, 2, 3, 4, 5, 6, 7, 8, 9],
[9, 8, 7, 6, 5, 4, 3, 2, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 2]
]
nums = m_list[1]
length = len(nums)
print(nums)
count_swap = 0
count_iter = 0
for i in range(length):
flag = False
for j in range(length -i -1):
count_iter += 1
if nums[j] > nums[j + 1]:
nums[j], nums[j + 1] = nums[j + 1], nums[j]
count_swap +=1
flag = True
if not flag:
break
print(nums, count_swap, count_iter)
5. 冒泡法总结
- 冒泡法需要数据一轮轮比较
- 可以设定一个标记判断此轮是否有数据交换发生,如果没有发生交换,可以结束排序,如果发生交换,继续下一轮排序
- 最差的排序情况是,初始顺序与目标顺序完全相反,遍历次数1,…,n-1之和n(n-1)/2
- 最好的排序情况是,初始顺序与目标顺序完全相同,遍历次数n-1
- 时间复杂度O(n2)