给定一个整数N ,任务是找到最接近N 且只有奇数位的数字。如果存在多个这样的数字,则打印最小值。
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
朴素的方法:解决问题的最简单的方法是迭代并找到最接近的小于和大于 N 的数字,只有奇数作为其数字。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。
- 找到最接近的更小和更大的数字,并返回与N差异最小的数字。如果发现两者与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: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与N最接近的较小数字:
- 从左到右求N中第一个偶数的位置,即pos 。
- 如果第一个偶数为0 ,则将0替换为9并迭代所有前面的数字,对于每个数字,如果发现为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 。
- 如果第一个偶数不为零,则将该数字减 1。用9替换所有后续数字。
- 通过以下步骤计算最接近N的较大数字:
- 找到第一个偶数的位置,即pos 。
- 将pos处的数字加 1。
- 迭代后续数字并将它们递增1。
- 比较与N获得的各个最接近数字的绝对差值,并打印出差值最小的那个。
- 如果发现两者的差异相等,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))