【leetcode】1387. Sort Integers by The Power Value

题目如下：

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 k-th 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 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 = 1, hi = 1, k = 1
Output: 1

Example 3:

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.

Example 4:

Input: lo = 10, hi = 20, k = 5
Output: 13

Example 5:

Input: lo = 1, hi = 1000, k = 777
Output: 570

Constraints:

• 1 <= lo <= hi <= 1000
• 1 <= k <= hi - lo + 1

解题思路：lo和hi的范围是1~1000，之间把范围内所有数字的power都算出来排序就好了。

代码如下：

class Solution(object):
    def getKth(self, lo, hi, k):
        """
        :type lo: int
        :type hi: int
        :type k: int
        :rtype: int
        """
        pair = []
        def transform(val):
            count = 0
            while val != 1:
                val = val / 2 if val % 2 == 0 else 3 * val + 1
                count += 1
            return count

        for i in range(lo,hi+1):
            pair.append((i,transform(i)))

        def cmpf(v1,v2):
            if v1[1] != v2[1]:
                return v1[1] - v2[1]
            return v1[0] - v2[0]

        pair.sort(cmpf)

        return pair[k-1][0]