【Leetcode】397. Integer Replacement

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本文探讨了LeetCode上的一道经典题目——整数替换问题。该题要求找到将任意正整数通过除以2（如果为偶数）、加1或减1（如果为奇数）的操作序列转换为1所需的最少步骤数。文章详细介绍了如何优化选择加1或减1操作以减少步骤的方法，并提供了一段实现这一策略的Java代码。

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题目链接：https://leetcode.com/contest/4/problems/integer-replacement/

题目：

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

UPDATE:
We have fixed the error in Integer Replacement. For input n = 3, the answer should be 2, not 3. Please try submitting again, thanks. We will rejudge your submissions after the contest finished and adjust your ranking accordingly.

思路：

easy 每遇到奇数时，分别判断n-1还是n+1的尾部零更多，越多的当然步骤越少。

特殊case：当n==Integer 最大值时，当n==3时

算法：

	public int integerReplacement(int n) {
		if (n == Integer.MAX_VALUE)
			return 32;
		int count = 0;
		while (n != 1) {
			if ((n & 1) == 0) {
				n = n >> 1;
				count++;
			} else {
				if (n == 3)
					n = 2;
				else {
					if (countTrailZero(n - 1) > countTrailZero(n + 1)) {
						n = n - 1;
					} else {
						n = n + 1;
					}
				}
				count++;
			}
		}
		return count;

	}

	public int countTrailZero(int n) {
		int c = 0;
		while ((n & 1) == 0) {
			n = n >> 1;
			c++;
		}
		return c;
	}