資訊專欄INFORMATION COLUMN
摘要：記一種簡(jiǎn)單的的做法先討論邊界，若為最大值，返回然后對(duì)整數(shù)分奇偶兩種情況討論，偶數(shù)除以，奇數(shù)判斷是否后能被整除且不等于，若如此則，否則每次操作后計(jì)數(shù)器，循環(huán)結(jié)束后返回計(jì)數(shù)器值。
Problem
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
記一種簡(jiǎn)單的iteration的做法：
先討論邊界case，若n為Integer最大值，返回32.
然后對(duì)整數(shù)n分奇偶兩種情況討論，偶數(shù)除以2，奇數(shù)判斷是否+1后能被4整除且n不等于3，若如此則+1，否則-1. 每次操作后計(jì)數(shù)器+1，循環(huán)結(jié)束后返回計(jì)數(shù)器值。
public class Solution {
public int integerReplacement(int n) {
if (n == Integer.MAX_VALUE) return 32;
int count = 0;
while (n != 1) {
if (n%2 == 0) n/=2;
else {
if ((n+1)%4==0 && n!=3) n+=1;
else n-=1;
}
count++;
}
return count;
}
}
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