描述
We have n chips, where the position of the ith chip is position[i].
We need to move all the chips to the same position. In one step, we can change the position of the ith chip from position[i] to:
- position[i] + 2 or position[i] - 2 with cost = 0.
- position[i] + 1 or position[i] - 1 with cost = 1.
Return the minimum cost needed to move all the chips to the same position.
Example 1:
Input: position = [1,2,3]
Output: 1
Explanation: First step: Move the chip at position 3 to position 1 with cost = 0.
Second step: Move the chip at position 2 to position 1 with cost = 1.
Total cost is 1.
Example 2:
Input: position = [2,2,2,3,3]
Output: 2
Explanation: We can move the two chips at position 3 to position 2. Each move has cost = 1. The total cost = 2.
Example 3:
Input: position = [1,1000000000]
Output: 1
Note:
1 <= position.length <= 100
1 <= position[i] <= 10^9
解析
根据题意和栗子可以看出来，不管多远的距离，只要 position 中的数字相差的距离是 2 的倍数消耗都是 0 ，数字相差的距离无法对 2 取余则消耗为 1 。position 中只有两类数字，一类是偶数，一类是奇数。如果只是一类（偶数或者奇数），那么不管将所有的分片移动到某个位置上的操作消耗都始终为 0 。而如果同时有奇数和偶数，最后的消耗就是 min(偶数个数，奇数个数) 。
解答
class Solution(object):
def minCostToMoveChips(self, position):
"""
:type position: List[int]
:rtype: int
"""
odd = 0
even = 0
for num in position:
if num % 2 == 0:
even += 1
else:
odd += 1
return min(even, odd)
运行结果
Runtime: 24 ms, faster than 43.78% of Python online submissions for Minimum Cost to Move Chips to The Same Position.
Memory Usage: 13.4 MB, less than 75.62% of Python online submissions for Minimum Cost to Move Chips to The Same Position.
原题链接：leetcode.com/problems/mi…
您的支持是我最大的动力
