바위타는 두루미
[leetcode]75. Sort Colors 본문
728x90
75. Sort Colors
Given an array with n objects colored red, white or blue, sort them in-place so that objects of the same color are adjacent, with the colors in the order red, white and blue.
Here, we will use the integers 0, 1, and 2 to represent the color red, white, and blue respectively.
Note: You are not suppose to use the library's sort function for this problem.
Example:
Input: [2,0,2,1,1,0] Output: [0,0,1,1,2,2]
Follow up:
- A rather straight forward solution is a two-pass algorithm using counting sort.
First, iterate the array counting number of 0's, 1's, and 2's, then overwrite array with total number of 0's, then 1's and followed by 2's. - Could you come up with a one-pass algorithm using only constant space?
첫번째 solution:
two-pass algorithm using counting sort.
class Solution(object):
def sortColors(self, nums):
count = [0]*3
for n in nums:
count[n] += 1
index = 0
for i in range(len(nums)):
while count[index] == 0:
index +=1
nums[i] = index
count[index]-=1
두번째 solution:
one-pass algorithm using only constant space
class Solution(object):
def sortColors(self, nums):
red_point = 0
blue_point = len(nums)-1
for i in range(len(nums)):
while nums[i] != 1 :
if nums[i] == 0:
if i == red_point :
red_point +=1
break
else :
nums[red_point], nums[i] = nums[i], nums[red_point]
red_point +=1
elif nums[i]==2:
if i> blue_point :
return
nums[blue_point], nums[i] = nums[i], nums[blue_point]
blue_point -=1
세번째 solution:
두번째 솔루션을 더 간단하게!
def sortColors(self, nums):
l=len(nums)
left=0
right=l-1
i=0
while i<=right:
if nums[i]==0:
nums[i],nums[left]=nums[left],nums[i]
left+=1
elif nums[i]==2:
nums[i],nums[right]=nums[right],nums[i]
right-=1
i-=1
i+=1
'Study > Algorithm' 카테고리의 다른 글
[leetcode]347. Top K Frequent Elements (0) | 2019.08.10 |
---|---|
[leetcode]347. Top K Frequent Elements (0) | 2019.08.10 |
[leetcode]79. Word Search (0) | 2019.08.10 |
[leetcode]78. Subsets (0) | 2019.08.10 |
[leetcode]46. Permutations (0) | 2019.08.10 |
Comments