1 Star 0 Fork 0

yuhang2__2/LeetCode-Solutions

加入 Gitee
与超过 1200万 开发者一起发现、参与优秀开源项目,私有仓库也完全免费 :)
免费加入
文件
克隆/下载
two-city-scheduling.py 1.50 KB
一键复制 编辑 原始数据 按行查看 历史
kamyu 提交于 6年前 . Update two-city-scheduling.py
# Time: O(n) ~ O(n^2), O(n) on average.
# Space: O(1)
import random
# quick select solution
class Solution(object):
def twoCitySchedCost(self, costs):
"""
:type costs: List[List[int]]
:rtype: int
"""
def kthElement(nums, k, compare):
def PartitionAroundPivot(left, right, pivot_idx, nums, compare):
new_pivot_idx = left
nums[pivot_idx], nums[right] = nums[right], nums[pivot_idx]
for i in xrange(left, right):
if compare(nums[i], nums[right]):
nums[i], nums[new_pivot_idx] = nums[new_pivot_idx], nums[i]
new_pivot_idx += 1
nums[right], nums[new_pivot_idx] = nums[new_pivot_idx], nums[right]
return new_pivot_idx
left, right = 0, len(nums) - 1
while left <= right:
pivot_idx = random.randint(left, right)
new_pivot_idx = PartitionAroundPivot(left, right, pivot_idx, nums, compare)
if new_pivot_idx == k:
return
elif new_pivot_idx > k:
right = new_pivot_idx - 1
else: # new_pivot_idx < k.
left = new_pivot_idx + 1
kthElement(costs, len(costs)//2, lambda a, b: a[0]-a[1] < b[0]-b[1])
result = 0
for i in xrange(len(costs)):
result += costs[i][0] if i < len(costs)//2 else costs[i][1]
return result
Loading...
马建仓 AI 助手
尝试更多
代码解读
代码找茬
代码优化
1
https://gitee.com/yuhang2__2/LeetCode-Solutions.git
git@gitee.com:yuhang2__2/LeetCode-Solutions.git
yuhang2__2
LeetCode-Solutions
LeetCode-Solutions
master

搜索帮助