LeetCode 1387. Sort Integers by The Power Value

The power of an integer x is defined as the number of steps needed to transform x into 1 using the following steps:

• if x is even then x = x / 2

• if x is odd then x = 3 * x + 1

For example, the power of x = 3 is 7 because 3 needs 7 steps to become 1 (3 --> 10 --> 5 --> 16 --> 8 --> 4 --> 2 --> 1).

Given three integers lo, hi and k. 

The task is to sort all integers in the interval [lo, hi] by the power value in ascending order, 

if two or more integers have the same power value sort them by ascending order.

Return the kth integer in the range [lo, hi] sorted by the power value.

Notice that for any integer x (lo <= x <= hi) 

it is guaranteed that x will transform into 1 using these steps and that the power of x is will fit in a 32-bit signed integer.

Example 1:

Input: lo = 12, hi = 15, k = 2

Output: 13

Explanation: The power of 12 is 9 (12 --> 6 --> 3 --> 10 --> 5 --> 16 --> 8 --> 4 --> 2 --> 1) The power of 13 is 9 

The power of 14 is 17 

The power of 15 is 17 

The interval sorted by the power value [12,13,14,15]. 

For k = 2 answer is the second element which is 13. 

Notice that 12 and 13 have the same power value and we sorted them in ascending order. Same for 14 and 15.

Example 2:

Input: lo = 7, hi = 11, k = 4

Output: 7

Explanation: 

The power array corresponding to the interval [7, 8, 9, 10, 11] is [16, 3, 19, 6, 14]. 

The interval sorted by power is [8, 10, 11, 7, 9]. The fourth number in the sorted array is 7.

Constraints:

• 1 <= lo <= hi <= 1000

• 1 <= k <= hi - lo + 1

之前是日本人提出的一个猜想，所有的数字如果是偶数就除以2，如果是奇数就乘以3+1，最后都会回归到1的。

1：先写个函数，判断这个数的power值是多少。

2：然后将lo到hi的数字放到二维数组中，第2个元素就是这个数字的power值；

3：对2维数组排序。

返回第k-1的数字即可；

Runtime: 62 ms, faster than 54.26% of Java online submissions for Sort Integers by The Power Value.

Memory Usage: 41.9 MB, less than 84.75% of Java online submissions for Sort Integers by The Power Value.