754. Reach a Number

LWC 65: 754. Reach a Number

传送门：754. Reach a Number

Problem

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： 本人没有特别好的思路，只能多举几个例子发现模式。例如，从0开始，第一步能够产生{+1, -1}， 第二步能够产生{+1, -1, +3, -3}，我们枚举到第六步，有：

0
1
1 3
0 2 4 6
0 2 4 6 8 10
1 3 5 7 9 11 13 15
1 3 5 7 9 11 13 15 17 19 21

注意：每一个数字代表的是绝对值，也就是说正负都能取到。观察最后一列数字，有：

0
1
3
6
10
15
21

对应的有：
1 = 0 + 1
3 = 0 + 1 + 2 = 3
6 = 0 + 1 + 2 + 3 = 6
...
21 = 0 + 1 + 2 + 3 + 4 + 5 + 6 = 21

所以首先我们要确定step，最后一个数用sum来表示，如果target > sum说明还没加够，必须让sum >= target， 但发现每两行是同奇或同偶，所以如果sum - target为偶数，说明在此行，直接输出，如果为奇数，那么step必须还得往后数，此时，step为偶数时 + 1， step为奇数时 + 2.

Java版本：

    public int reachNumber(int target) {
        target = Math.abs(target);
        int sum = 0;
        int n   = 0;
        while (sum < target){
            n ++;
            sum += n;
        }
        if (sum == target) return n;
        int res = sum - target;
        if ((res & 1) == 0) {
            return n;
        }
        else {
            return n + ((n & 1) == 0 ? 1 : 2);
        }
    }

Python版本：

class Solution(object):
    def reachNumber(self, target):
        """
        :type target: int
        :rtype: int
        """
        t = abs(target)
        n = 0
        sum = 0
        while (t > sum):
            n += 1
            sum += n
        if t == sum: return n
        rem = sum - target
        if (rem % 2) == 0: return n
        else:
            return n + 1 if (n % 2) == 0 else n + 2