【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 transformx
into1
using the following steps:
- if
x
is even thenx = x / 2
- if
x
is odd thenx = 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
andk
. 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 thatx
will transform into1
using these steps and that the power ofx
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: 1Example 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: 13Example 5:
Input: lo = 1, hi = 1000, k = 777 Output: 570Constraints:
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]