【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;
    }
};
 
 
 

 

posted @ 2019-01-05 20:14  wikiwen  阅读(222)  评论(0编辑  收藏