【LeetCode & 剑指offer刷题】查找与排序题11:Sort Colors

【LeetCode & 剑指offer 刷题笔记】目录(持续更新中...)

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?

C++
 
//问题:颜色排序(以后看可不可以用partition)
//双指针法,只能用于只有三类的排序,不过只需一次遍历
#include <algorithm>
class Solution
{
public :
    void sortColors ( vector < int >& A )
    {
        int left = 0 , right = A . size ()- 1 ; // 双指针, left 从左边扫描, right 从右边扫描
        for ( int i = 0 ; i <= right ; i ++) // 扫描到 right 即可
        {
            if ( A [ i ] == 0 ) swap ( A [ i ++], A [ left ++]); // 0 换到左边
            else if ( A [ i ] == 2 ) swap ( A [ i --], A [ right --]); // 2 换到右边, i-- 是以免 right 换过来的值为 0(抵消for循环中的i++)
        }
    }
};
//存储后移法(相当于直接插入排序,只不过预先知道了有哪些数,故简单一点)
class Solution
{
public :
    void sortColors ( vector < int >& A )
    {
        int n0 , n1 , n2 ; //类别索引
        n0 = n1 = n2 = 0 ;
        for ( int i = 0 ; i < A . size (); i ++)
        {
            switch ( A [ i ])
            {
                case 0 :
                    A [ n2 ++] = 2 ; A [ n1 ++] = 1 ; A [ n0 ++] = 0 ; //后面的元素往后移
                    break ;
                case 1 :
                    A [ n2 ++] = 2 ; A [ n1 ++] = 1 ;
                    break ;
                case 2 :
                    A [ n2 ++] = 2 ;
            }
        }
    }
};
 
// 计数排序,需要两次遍历
class Solution
{
public :
    void sortColors ( vector < int >& nums )
    {
        int count [ 3 ] = { 0 }; // 用于统计每个颜色出现的次数
        for ( int i = 0 ; i < nums . size (); i ++) // 统计直方图
            count [ nums [ i ]]++;
      
        for ( int i = 0 , index = 0 ; i<3 ; i ++) // 排序
            for ( int j = 0 ; j < count[i ]; j ++)
                nums [ index ++] = i ;
    }
};
 
 
 

 

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转载自www.cnblogs.com/wikiwen/p/10225956.html
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