397.leetcode Integer Replacement(easy)[数字处理 溢出]

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

解法一采用传统的函数递归处理

解法二通过观察发现除了3以外其他+1是4的倍数的数都采用加一得到的次数更小

class Solution {
public:
解法一:递归迭代

int getResult(int n)
{
if(n == 1) return 0;
int count = 0;
if(n%2 == 0)
{
n = n>>1;
++count;
count += getResult(n);

}else
{
++count;
long long t = n;
int a = getResult((t-1)/2);
int b = getResult((t+1)/2);
++count;
count += a>b?b:a;

}
return count;
}
int integerReplacement(int n) {
if(n<=1) return 0;
return getResult(n);
}

解法二总结算法
int integerReplacement(int n) {
long long t = n;
int count = 0;
while (t > 1) {
++count;
if (t & 1) {
if ((t & 2) && (t != 3)) ++t;
else --t;
} else {
t >>= 1;
}
}
return count;
}
};