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LeetCode 754. Reach a Number

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题目

题目链接
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].

分析

  1. target的正负不影响计算结果,因为数轴是对称的target = abs(target).
  2. 找到整数n,使得$(\sum_{i=1}^n i) \geq target$.
    1. $有求和公式: target = \frac{n*(n+1)}{2}$.
    2. $整形得到一元二次方程: n^2+n-2*target=0$.
    3. $公式求解: n = \frac{(-1 + \sqrt{(1 + 8*target)})}{2}$.
    4. n可能不为整,向上取整即可.
  3. 取$sum = \frac{n*(n+1)}{2}$, sum 与 target 的差 diff = sum - target
  4. 此时,如果 sum == target,直接返回n即可. 又或者差值diff是偶数,那么只要把$\frac{\text{diff}}{2}$取反即可得到target,所以也是n.
  5. 若差值diff为奇数,则有如下两种情况:
    1. n为奇数,则n+1为偶数,需要$\sum_{i=1}^{n+2} i$ 来改变diff的奇偶.
    2. n为偶数,则n+1为奇数,$\sum_{i=1}^{n+1} i$ 来改变diff的奇偶.

代码

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class Solution {
public:
    int reachNumber(int target) {
        long long t = abs(target);
        long long n = ceil((-1 + sqrt(1 + 8*t)) / 2);
        long long sum = (n * (n + 1)) / 2;
        long long diff = sum - t;
        if (sum == target || !(diff % 2)) return n;
        return n + ((n % 2) ? 2 : 1);
    }
};