作者： 負雪明燭
id： fuxuemingzhu
個人部落格： http://fuxuemingzhu.cn/

目錄

• 題目描述
• 題目大意
• 解題方法
• 數學
• 日期

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

題目描述

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].

題目大意

每次走的步數是增加1步，方向是可以向左或者向右，求通過多少步之後能到達target。

解題方法

數學

非常不喜歡數學題，所以花花醬和Grandyang大神的帖子粘在這裡了。

花花醬：https://zxi.mytechroad.com/blog/math/leetcode-754-reach-a-number/
Grandyang大神：http://www.cnblogs.com/grandyang/p/8456022.html

class Solution(object):
    def reachNumber(self, target):
        """
        :type target: int
        :rtype: int
        """
        target = abs(target)
        k = 0
        sum = 0
        while sum < target:
            k += 1
            sum += k
        d = sum - target
        if d % 2 == 0:
            return k
        return k + 1 + (k % 2)

C++版本如下：

class Solution {
public:
    int reachNumber(int target) {
        target = abs(target);
        int k = 0;
        int sum = 0;
        while (sum < target) {
            sum += (++k);
        }
        const int d = sum - target;
        if (d % 2 == 0) return k;
        return k + 1 + (k % 2);
    }
};

日期

2018 年 11 月 26 日 —— 11月最後一週！