[Leetcode] 754. Reach a Number 解题报告

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].

思路：

我开始用BFS实现了，但是TLE了。如果把移动n步所可以达到的位置记录下来，则我们可以得到如下的表格：

[0]
[-1, 1]
[-3, 1, -1, 3]
[-6, 0, -2, 4, -4, 2, 0, 6]
[-10, -2, -4, 4, -6, 2, 0, 8, -8, 0, -2, 6, -4, 4, 2, 10]
[-15, -5, -7, 3, -9, 1, -1, 9, -11, -1, -3, 7, -5, 5, 3, 13, -13, -3, -5, 5, -7, 3, 1, 11, -9, 1, -1, 9, -3, 7, 5, 15]
[-21, -9, -11, 1, -13, -1, -3, 9, -15, -3, -5, 7, -7, 5, 3, 15, -17, -5, -7, 5, -9, 3, 1, 13, -11, 1, -1, 11, -3, 9, 7, 19, -19, -7, -9, 3, -11, 1, -1, 11, -13, -1, -3, 9, -5, 7, 5, 17, -15, -3, -5, 7, -7, 5, 3, 15, -9, 3, 1, 13, -1, 11, 9, 21]
[-28, -14, -16, -2, -18, -4, -6, 8, -20, -6, -8, 6, -10, 4, 2, 16, -22, -8, -10, 4, -12, 2, 0, 14, -14, 0, -2, 12, -4, 10, 8, 22, -24, -10, -12, 2, -14, 0, -2, 12, -16, -2, -4, 10, -6, 8, 6, 20, -18, -4, -6, 8, -8, 6, 4, 18, -10, 4, 2, 16, 0, 14, 12, 26, -26, -12, -14, 0, -16, -2, -4, 10, -18, -4, -6, 8, -8, 6, 4, 18, -20, -6, -8, 6, -10, 4, 2, 16, -12, 2, 0, 14, -2, 12, 10, 24, -22, -8, -10, 4, -12, 2, 0, 14, -14, 0, -2, 12, -4, 10, 8, 22, -16, -2, -4, 10, -6, 8, 6, 20, -8, 6, 4, 18, 2, 16, 14, 28]

这个还不够清楚。如果我们将每行都进行排序，则可以得到如下结果：

[0]
[-1, 1]
[-3, -1, 1, 3]
[-6, -4, -2, 0, 2, 4, 6]
[-10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10]
[-15, -13, -11, -9, -7, -5, -3, -1, 1, 3, 5, 7, 9, 11, 13, 15]
[-21, -19, -17, -15, -13, -11, -9, -7, -5, -3, -1, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21]
[-28, -26, -24, -22, -20, -18, -16, -14, -12, -10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28]
[-36, -34, -32, -30, -28, -26, -24, -22, -20, -18, -16, -14, -12, -10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36]
[-45, -43, -41, -39, -37, -35, -33, -31, -29, -27, -25, -23, -21, -19, -17, -15, -13, -11, -9, -7, -5, -3, -1, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45]

发现什么规律了吗？总结起来就是：

1）在每一行中，如果target在该行中，则-target也一定在该行中。

2）每一行的中的所有数的奇偶性相同。

3）第n行的最大值（最小值）就是连续向右（左）走所可以达到的位置，最大值就是n * (n + 1) / 2，最小值就是-n * (n + 1) / 2。

4）如果target介于某一行的最大值和最小值之间，其奇偶性和该行数的奇偶性相同，那么target一定在这一行中。

有了以上观察，则思路就清楚了：我们一直累加sum直到sum >= target并且他们两个的奇偶性相同。那么此时target一定在step所在的行，所以返回step就可以了。

代码：

class Solution {
public:
    int reachNumber(int target) {
        target = abs(target);
        int step = 0, sum = 0;
        while (sum < target || (sum - target) % 2 != 0) {
            sum += (++step);
        }
        return step;
    }
};