给定整数N ，任务是找到最接近N的仅具有奇数位的数字。如果存在多个这样的数字，请打印最小值。
例子：
Input: N = 725
Output: 719
Explanation:
Numbers 719 and 731 consists of odd digits only and are closest to 725. Since 719 is the minimum, it is the required answer.
Input : N = 111
Output: 111
天真的方法：解决问题的最简单方法是迭代并找到小于和大于N的最接近的数字，仅将奇数作为其数字。请按照以下步骤解决问题：
- 如果N的所有数字都是奇数，则打印N作为答案。
- 找到最接近的较小和较大的数字，然后返回与N最小的数字。如果发现两者都与N等距，则打印较小的数字。
时间复杂度：O(10 (len(N)– 1) )
辅助空间： O(1)
高效方法：请按照以下步骤优化上述方法：
- 如果N的所有数字都是奇数，则返回N。
- 通过以下步骤有效地计算最接近N的较小数字：
- 从左到右找到N中第一个偶数位的位置，即pos 。
- 如果第一个偶数为0 ，则将0替换为9 ，然后遍历所有前面的数字；对于每个数字，如果找到的是1 ，则将其替换为9 。否则，将数字减2并返回数字。
- 如果没有发现前面的数字超过1 ，则将最高有效数字替换为0 。
- 如果第一个偶数数字不为零，则将该数字减1。将所有后续数字替换为9 。
- 通过以下步骤计算最接近N的较大数：
- 找到第一个偶数数字的位置，即pos 。
- 将pos处的数字加1。
- 遍历后续数字并将其递增1。
- 将获得的各个最接近的数字的绝对差与N进行比较，并打印出差异最小的数字。
- 如果发现两个差异相等，则打印较小的数字。
下面是上述方法的实现：
Python
# Python program to implement
# the above approach
# Function to return the smaller
# number closest to N made up of
# only odd digits
def closest_smaller(N):
N = str(N)
l = list(map(int, N))
length = len(l)
# Stores the position of
# first even digit of N
pos = -1
# Iterate through each digit of N
for i in range(length):
# Check for even digit.
if l[i] % 2 == 0:
pos = i
break
# If the first even digit is 0
if l[pos] == 0:
# Replace 0 with 9
l[pos] = 9
# Iterate over preceding
# digits
for i in range(pos - 1, -1, -1):
# If current digit is 1
if l[i] == 1:
# Check if it is the
# first digit or not
if i == 0:
# Append leading 0's
l[i] = 0
break
# Otherwise, replace by 9
l[i] = 9
# Otherwise
else:
# Decrease its value by 2
l[i] -= 2
break
# If the first even digit exceeds 0
else:
# Reduce the digit by 1
l[pos] -= 1
# Replace all succeeding digits by 9
for i in range(pos + 1, length):
l[i] = 9
# Remove leading 0s
if l[0] == 0:
l.pop(0)
result = ''.join(map(str, l))
return result
# Function to return the greater
# number closest to N made up of
# only odd digits
def closest_greater(N):
N = str(N)
l = list(map(int, N))
length = len(l)
# Stores the position of
# first even digit of N
pos = -1
# Iterate over each digit
# of N
for i in range(length):
# If even digit is found
if l[i] % 2 == 0:
pos = i
break
# Increase value of first
# even digit by 1
l[pos] += 1
for i in range(pos + 1, length):
l[i] = 1
result = ''.join(map(str, l))
return result
# Function to check if all
# digits of N are odd or not
def check_all_digits_odd(N):
N = str(N)
l = list(map(int, N))
length = len(l)
# Stores the position of
# first even digit of N
pos = -1
# Iterating over each digit
# of N
for i in range(length):
# If even digit is found
if l[i] % 2 == 0:
pos = i
break
# If no even digit is found
if pos == -1:
return True
return False
# Function to return the
# closest number to N
# having odd digits only
def closestNumber(N):
# If all digits of N are odd
if check_all_digits_odd(N):
print(N)
else:
# Find smaller number
# closest to N
l = int(closest_smaller(N))
# Find greater number
# closest to N
r = int(closest_greater(N))
# Print the number with least
# absolute difference
if abs(N - l) <= abs(N - r):
print(l)
else:
print(r)
# Driver Code.
if __name__ == '__main__':
N = 110
closestNumber(N)输出：
111
时间复杂度： O(len(N))
辅助空间： O(len(N))