0397._Integer_Replacement.md

397. Integer Replacement

难度: Medium

刷题内容

原题连接

• https://leetcode.com/problems/integer-replacement/

内容描述

Given a positive integer n and you can do operations as follow:

If n is even, replace n with n/2.
If n is odd, you can replace n with either n + 1 or n - 1.
What is the minimum number of replacements needed for n to become 1?

Example 1:

Input:
8

Output:
3

Explanation:
8 -> 4 -> 2 -> 1
Example 2:

Input:
7

Output:
4

Explanation:
7 -> 8 -> 4 -> 2 -> 1
or
7 -> 6 -> 3 -> 2 -> 1

解题方案

思路 1 - 时间复杂度: O(lgN)- 空间复杂度: O(1)******

无脑递归, beats 15.63%

class Solution:
    def integerReplacement(self, n):
        """
        :type n: int
        :rtype: int
        """
        if n == 1:
            return 0
        if n % 2 == 0:
            return 1 + self.integerReplacement(n // 2)
        return 1 + min(self.integerReplacement(n - 1), self.integerReplacement(n + 1))

思路 2 - 时间复杂度: O(lgN)- 空间复杂度: O(1)******

烧脑数学证明

beats 59.38%

class Solution:
    def integerReplacement(self, n):
        """
        :type n: int
        :rtype: int
        """
        res = 0
        while n > 1:
            res += 1
            if n % 2 == 0:
                n //= 2
            elif n % 4 == 1 or n == 3:
                n -= 1
            else:
                n += 1
        return res

one explantion is intuitive (@shshwdr from wechat):

n = 2k + 1
n + 1 = 2k+2         (n+1) / 2 = k+1
n - 1 = 2k           (n-1) / 2 = k
one of k and k+1 will be even, we want to get even number optimally at the end, 
so we check which one is multiplier of 4 and choose it

but notice one edge case, n = 3, we would choose n - 1 = 2 instead of n + 1 = 4