哈利波特获得了一个数字列表，他的任务是必须将奇数按升序排序，同时将偶数保留在原来的位置。魔法师们，快来帮帮他吧～
题目难度：中等
题目来源：codewars
def sort_the_odd(nums):
pass
assert sort_the_odd([7, 1]) == [1, 7]
assert sort_the_odd([5, 8, 6, 3, 4]) == [3, 8, 6, 5, 4]
assert sort_the_odd([9, 8, 7, 6, 5, 4, 3, 2, 1, 0]) == [1, 8, 3, 6, 5, 4, 7, 2, 9, 0]
def sort_the_odd(nums:list=None):
""""""
# print(nums)
odd_number= sorted([n for n in nums if n%2!=0]) #过滤出奇数并排序
for index in range(len(nums)):
if nums[index]%2 != 0:#奇数
nums[index] = odd_number[0] #将源列表中的奇数依次替换掉排好序的奇数
odd_number.pop(0)
return nums
def sort_the_odd2(nums:list=None):
even_num = [(nums.index(num),num) for num in nums if num%2 == 0] #拿源列表偶数的位置和值
odd_number= sorted([n for n in nums if n%2!=0]) #过滤出奇数并排序
for index,num in even_num:
odd_number.insert(index,num) #在奇数队列中插入偶数
return odd_number
冒泡排序的升级版吗？ 求简单解法。。。
def sort_the_odd(li):
li_length=len(li)
li1=[]
li2=[]
for i in range(li_length):
if li[i]%2==1:
li1.append(i)
li2.append(li[i])
for i in range(len(li2)-1):
for j in range(len(li2)-1-i):
if li2[j]>li2[j+1]:
li2[j],li2[j+1]=li2[j+1],li2[j]
for i in range(len(li2)):
li[li1[i]]=li2[i]
return li
def sort_the_odd(nums):
odds = sorted(filter(lambda x:x % 2 == 1, nums))
return [odds.pop(0) if item % 2 == 1 else item for i,item in enumerate(nums)]
assert sort_the_odd([7, 1]) == [1, 7]
assert sort_the_odd([5, 8, 6, 3, 4]) == [3, 8, 6, 5, 4]
assert sort_the_odd([9, 8, 7, 6, 5, 4, 3, 2, 1, 0]) == [1, 8, 3, 6, 5, 4, 7, 2, 9, 0]
for (int i = 0; i < nums.length; i++) {
if (nums[i] % 2 == 0) {
continue;
} else {
for (int j = i; j < nums.length; j++) {
if ((nums[i] > nums[j]) &&(nums[j]%2 !=0)) {
int tmp = nums[i];
nums[i] = nums[j];
nums[j] = tmp;
}
}
}
}