zoukankan      html  css  js  c++  java
  • [leetcode sort]75. Sort Colors

    Given an array with n objects colored red, white or blue, sort them 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.

    解法1:

    统计每个颜色出现的次数,这个方法比较简单易懂

     1 class Solution(object):
     2     def sortColors(self, nums):
     3         flag = [0,0,0]
     4         for i in nums:
     5             flag[i] += 1
     6         for n in range(flag[0]):
     7             nums[n] = 0
     8         for n in range(flag[1]):
     9             nums[flag[0]+n]=1
    10         for n in range(flag[2]):
    11             nums[flag[0]+flag[1]+n] = 2

    解法2:

    在评论区总是能发现一些很漂亮的方法,方法和快排比较相似

     1 class Solution(object):
     2     def sortColors(self, nums):
     3         """
     4         :type nums: List[int]
     5         :rtype: void Do not return anything, modify nums in-place instead.
     6         """
     7         i,start,end = 0,0,len(nums)-1
     8         while i<=end:
     9             if nums[i]==0:
    10                 nums[i]=nums[start]
    11                 nums[start]=0
    12                 start += 1
    13             elif nums[i]==2:
    14                 nums[i]=nums[end]
    15                 nums[end]=2
    16                 end -= 1
    17                 i -= 1 #交换后此位置的数未考虑,要重新考虑
    18             i += 1

    解法3:

    最帅的是这种方法,类似于快排,指针i,j分别处理以类数据

     1 class Solution(object):
     2     def sortColors(self, nums):
     3         """
     4         :type nums: List[int]
     5         :rtype: void Do not return anything, modify nums in-place instead.
     6         """
     7         i,j = 0,0
     8         for k in range(len(nums)):
     9             v = nums[k]
    10             nums[k] = 2
    11             if v<2:
    12                 nums[j]=1
    13                 j += 1
    14             if v == 0:
    15                 nums[i]=0
    16                 i += 1
  • 相关阅读:
    Acwing 164 可达性统计 (拓扑排序+bitset)
    STL-bitset的基本用法
    Acwing 115 给树染色 (贪心)
    Acwing 112 雷达设备 (贪心)
    Acwing 110 畜栏预定 (贪心+stl)
    Acwing 110 防晒 (贪心算法)
    Acwing 七夕祭 (前缀和+中位数+思维)
    Acwing 103 电影 (map)
    USACO 最佳牛围栏 (二分答案+前缀和+双指针)
    Acwing 101 最高的牛 (差分数组)
  • 原文地址:https://www.cnblogs.com/fcyworld/p/6511535.html
Copyright © 2011-2022 走看看