754. Reach a Number

问题

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].

解决方法

对于target而言，其正负由于对称性，所走的步数是一样的。比如，-1和+1都是只走一步，区别是往左还是往右而已。所以该问题可以简化为只考虑target为正的情况。
现在我们来考虑如何分步解决该问题。如果要得到最小的步数，必须尽量一直往右走，任何的后退都会让步数增多。假定存在k，S=1+2+...+k，有如下几种情况：

1. S==target，则k为最小步数。
2. S
3. S>target，需要往左走才能到达target。具体该如何走我们来详细讨论。

往左走的步数越少越好，能否往左只走一步呢？假定第n步往左走（1<=n<=k）,一共会走1+2+...+n-1 - n + n+1+...+k=(1+2+...+n-1 + n + n+1 + ... + k) - 2n= S-2n=target，即S-target=2n。也就是说，如果S与target之间的差值为偶数，才能保证调整n的方向，就可以在k步走到target。如果差值为奇数，我们需要继续往前走一步，使差值变为偶数。

class Solution:
    def reachNumber(self, target: int) -> int:
        target_ = abs(target)
        sum_, step = 0, 0
        
        while sum_ < target_ or (sum_ - target_) & 1 != 0:
            step += 1
            sum_ += step
            
        return step

参考

https://www.cnblogs.com/grandyang/p/8456022.html
https://www.geeksforgeeks.org/minimum-steps-to-reach-a-destination/