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:
Input: 7
Output: 4
Explanation: 7 -> 8 -> 4 -> 2 -> 1 or 7 -> 6 -> 3 -> 2 -> 1
Tips: The infamous test with n=3 fails for that strategy because 11 ->10 -> 1 is better than 11 -> 100 -> 10 -> 1.
Solution:
If n is even, halve it.
If n=3 or n-1 has less 1"s than n+1,
decrement n. Otherwise, increment n.
Code:
public int integerReplacement(int n){ int count = 0; while(n!= 1){ if((n&1)==0){ n>>>=1; }else if(n ==3 ||((n >>>1) & 1 ==0){ --n; } else{ ++n; } count++; } return count; }
文章版权归作者所有,未经允许请勿转载,若此文章存在违规行为,您可以联系管理员删除。
转载请注明本文地址:https://www.ucloud.cn/yun/66429.html
摘要:复杂度思路利用位的操作。如果一个数是奇数,那么末位的位一定是。对于偶数,操作是直接除以。对于奇数的操作如果倒数第二位是,那么的操作比的操作能消掉更多的。还有一个的地方是,为了防止越界,可以将先转换成。 LeetCode[397] Integer Replacement Given a positive integer n and you can do operations as fo...
摘要:题目要求思路和代码可以发现除二后所得到的结果一定优于加减。因此,如果当前奇数除二为偶数,则直接做除法,否则将当前奇数加一再除以二,得到偶数的结果。 题目要求 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...
Problem Given a positive integer n and you can do operations as follow: 1.If n is even, replace n with n/2.2.If n is odd, you can replace n with either n + 1 or n - 1. What is the minimum number of re...
摘要:记一种简单的的做法先讨论边界,若为最大值,返回然后对整数分奇偶两种情况讨论,偶数除以,奇数判断是否后能被整除且不等于,若如此则,否则每次操作后计数器,循环结束后返回计数器值。 Problem Given a positive integer n and you can do operations as follow: If n is even, replace n with n/2.I...
摘要:解题思路这题就是最基础的递归运算题目,两个选择,一个是偶数,一个是奇数,偶数直接除操作,奇数变成左右两个偶数继续操作选择操作最小的,注意有一个用例是,解决方法有两种,第一就是首先把的二次幂都干掉,代码如下 ...
阅读 3513·2021-11-17 17:01
阅读 3918·2021-11-08 13:12
阅读 2477·2021-10-08 10:04
阅读 687·2021-09-29 09:35
阅读 1418·2021-09-26 10:12
阅读 2020·2021-09-07 09:58
阅读 1953·2019-08-30 15:55
阅读 2134·2019-08-30 13:14