【LeetCode】754. Reach a Number（C++）

地址：https://leetcode.com/problems/reach-a-number

题目：

You are standing at position 0 on an infinite number line. There is a goal at position target.

On each move, you can either go left or right. During the n-th move (starting from 1), you take n steps.

Return the minimum number of steps required to reach the destination.
Example 1:

Input: target = 3
Output: 2
Explanation:
On the first move we step from 0 to 1.
On the second step we step from 1 to 3.

Example 2:

Input: target = 2
Output: 3
Explanation:
On the first move we step from 0 to 1.
On the second move we step from 1 to -1.
On the third move we step from -1 to 2.

Note:

• target will be a non-zero integer in the range [-10^9, 10^9].

理解：

首先，若能达到target，则一定能通过相同的步骤达到-target。因为若
$$target=\sum _{i=1}^k{a}_{i}i$$
则
$$-target=\sum _{i=1}^k-a_{i}i$$
因此只需考虑正的target即可。

• 首先，寻找第一个大于target的sum，使得$$sum=1+2+\cdots +k>=target$$
• 若target==sum，则返回k
• 若diff=sum-target是偶数，则可以通过把$\frac{\text{sum}-\text{target}}{2}$的符号取反，从而得到target。
  解释：
  $$\text{sum}=1+\cdots +\frac{\text{sum}-\text{target}}{2}+\cdots +k$$
  $$\text{target}=1+\cdots -\frac{\text{sum}-\text{target}}{2}+\cdots +k$$
  观察上面的两个式子，因为$\text{sum}-\text{target}$是偶数，且$\text{sum}-\text{target}<k$，因此总可以在 { 1 ⋯ k } \left\{ 1\cdots k\right\} {1⋯k}中找到一个数$i=\frac{\text{sum}-\text{target}}{2}$，将其符号取反，就得到了$\text{target}$。
  也就是说，当$\text{sum}-\text{target}$为偶数的时候，通过反转$\frac{\text{sum}-\text{target}}{2}$的符号，可以把sum减小到target。
• 当$\text{sum}-\text{target}$为奇数的时候
  • 如果n是偶数，n+1是奇数，则加上n+1，sum-target就变成了偶数，然后进行翻转即可
  • 若n是偶数，n+1是偶数，则加上n+1和n+2，sum-target就变成了偶数，然后进行翻转即可

实现：

class Solution {
public:
	int reachNumber(int target) {
		target = abs(target);
		//8.0 is necessary, otherwise 8*target is overflow
		long long k = ceil((-1 + sqrt(1 + 8.0 * target)) / 2.0);
		long long sum = (1 + k)*k / 2;
		if (sum == target) return k;
		long long diff = sum - target;
		if ((diff & 1) == 0)
			return k;
		else
            //k is odd, k+1 is even, k+2 is needed
			return (k & 1) ? k + 2 : k + 1;
	}
};

总感觉理解的地方写得不够严谨，特别是总能找到的证明是不是有问题啊。。先不管了