• No.75 Sort Colors


    No.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.

    解法:

      颜色排序:数组nums中包括代表红白蓝三色的0/1/2的n个数,
      输出:经排序后,nums是有序的红白蓝即0/1/2有序排列
      注意:0、1、2连续,可为数组下标,更简洁

     1 class Solution
     2 {
     3 public:
     4     void sortColors(vector<int> &nums)
     5     {//颜色排序:数组nums中包括代表红白蓝三色的0/1/2的n个数,
     6      //输出:经排序后,nums是有序的红白蓝即0/1/2有序排列
     7      //注意:0、1、2连续,可为数组下标,更简洁
     8         int count = nums.size();
     9         if(count <=1)
    10             return;
    11         int color[3] = {0};//颜色统计数组
    12         
    13         for(int i=0; i<count; i++)
    14             color[nums[i]]++;//技巧
    15 
    16         for(int i=0,index=0; i<3; i++)
    17             for(int j=0; j<color[i]; j++)
    18                 nums[index++] = i;
    19         //so简洁,so easy
    20     }
    21 };

    我的笨代码:

     1 #include "stdafx.h"
     2 #include <vector>
     3 #include <iostream>
     4 using namespace std;
     5 
     6 class Solution
     7 {
     8 public:
     9     void sortColors(vector<int> &nums)
    10     {//颜色排序:数组nums中包括代表红白蓝三色的0/1/2的n个数,
    11      //输出:经排序后,nums是有序的红白蓝即0/1/2有序排列
    12         int count = nums.size();
    13         if(count <= 1)
    14             return;
    15 
    16         //类似计数排序
    17         int countRed = 0;
    18         int countWhite = 0;
    19         int countBlue = 0;
    20         //从前向后遍历
    21         for(int i=0; i<count; i++)
    22         {
    23             if(nums[i] == 0)
    24                 countRed++;
    25             else if(nums[i] == 1)
    26                 countWhite++;
    27             else if(nums[i] == 2)
    28                 countBlue++;
    29         }
    30 
    31         //根据统计结果,从前向后改变数组
    32         for(int i=0; i<count; i++)
    33         {
    34             if(countRed != 0)
    35             {
    36                 nums[i] = 0;
    37                 countRed--;
    38                 continue;
    39             }
    40             if(countWhite != 0)
    41             {
    42                 nums[i] = 1;
    43                 countWhite--;
    44                 continue;
    45             }
    46             if(countBlue != 0)
    47             {
    48                 nums[i] = 2;
    49                 countBlue--;
    50                 continue;
    51             }
    52         }
    53     }     
    54 };
    55 int main()
    56 {
    57     Solution sol;
    58     int data[] = {0,2,1,2,1,2,0,0,0,1};
    59     vector<int> test(data,data+sizeof(data)/sizeof(int));
    60 
    61     for(auto &i : test)
    62         cout << i << " ";
    63     cout << endl;
    64     sol.sortColors(test);
    65         for(auto &i : test)
    66         cout << i << " ";
    67     cout << endl;
    68 
    69     return 0;
    70 }
    View Code
  • 相关阅读:
    [转]实习生需要懂的40大基本规矩
    [转]Linux下pppoe配合Drcom插件上网方法介绍......
    收藏一些图书
    [转]30个自我提升技巧
    [转]关于Gmail打不开的解决办法
    [转]李开复经典语录盘点:人生之路在于每次的选择
    [转]哈佛管理世界中智慧
    胡伟武校友在2011年中国科大本科生毕业典礼暨学位授予仪式上的讲话
    Dynamics4.0和Dynamics2011处理Email的方法
    JS实现简单的ToolTip功能
  • 原文地址:https://www.cnblogs.com/dreamrun/p/4561300.html
Copyright © 2020-2023  润新知