leetcode75-- Sort Color

leetcode75-- Sort Color

题目

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?

解决

我先想到的是用数学的办法直接算出0,1,2的个数,

class Solution(object):
    def sortColors(self, nums):
        """
        :type nums: List[int]
        :rtype: None Do not return anything, modify nums in-place instead.
        """
        
        summ1 = 0
        summ2 = 0
        for x in nums:
            if x==1:
                summ1 += 1
                summ2 += 2
            elif x==2:
                summ1 += 2
                summ2 += 2
                
        nums[:] = [0]*(len(nums)-(summ2-summ1)-(2*summ1-summ2)/2)+[1]*(summ2-summ1)+[2]*((2*summ1-summ2)/2)
        # print(ret)
        # nums[:] = ret

summ1 = y+2z; summ2 = 2y+2z 其中y是1的个数,z是2的个数.

也可以利用投桶的方式

class Solution(object):
    def sortColors(self, nums):
        """
        :type nums: List[int]
        :rtype: None Do not return anything, modify nums in-place instead.
        """
        ret = [0]*3
        
        for x in nums:
            ret[x] += 1
            
        nums[:] = [0]*ret[0]+[1]*ret[1]+[2]*ret[2]
        

也可以利用双指针的办法

class Solution(object):
    def sortColors(self, nums):
        """
        :type nums: List[int]
        :rtype: None Do not return anything, modify nums in-place instead.
        """
        
        red, blue = 0, len(nums)-1
        i = 0 
        while i<= blue:
            if nums[i]==0:
                nums[i], nums[red] = nums[red], nums[i]
                red += 1
            elif nums[i]==2:
                nums[i], nums[blue] = nums[blue], nums[i]
                blue -= 1
                i -= 1  # 这里减去1的原因是blue的刚换过来一个,刚换过来的还没有处理.
          
            i += 1
        
                

即刚开始假设red=0,blue=len(nums)-1,然后开始遍历,见到0的话,就把这个数与nums[red]换,并且red 自加一次,见到2的时候就和nums[blue]对换,blue自减一次,见到1的时候不用动,因为1是中间值, 停止的条件是i<=blue, 即当前的位置不能超过blue, 因为当i与blue相见的时候,后面的全部都是2了,而且这时候0也排好了(因为i都走到现在了),所以这就意味着全部都处理完了. 前面的方法也是可行的,不过最后一种方法很巧妙,相当于看到了排完序的最终的状态,所以就只针对红的和蓝的进行处理,而且用的空间比较少.

打赏,谢谢~~

取消

感谢您的支持,我会继续努力的!

扫码支持
扫码打赏,多谢支持~

打开微信扫一扫,即可进行扫码打赏哦