【题目】
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.
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 an one-pass algorithm using only constant space?
【题意】
对只有三种值0、1、2的数组进行排序,要求扫描一遍就排好序
【思路】
排好序之后0在前,1在中,2在后
维护两个p1和p2,p1指向0数组的尾部,p2指向2数组头部
在维护一个p指针用于扫描数组,当p指向0时与p1置换,当p指向12时与p2置换
【注意,在给定的数组中,3个值不一定都出现】
【代码】
class Solution { public: void sortColors(int A[], int n) { if(n==0 || n==1)return; int p1=0, p2=n-1; while(A[p1]==0)p1++; while(A[p2]==2)p2--; int p=p1; while(p<=p2){ if(A[p]==0){ if(p==p1){p1++; p++;} else{ A[p]=A[p1]; A[p1]=0; while(A[p1]==0)p1++; } } else if(A[p]==2){ if(p==p2)p++; else{ A[p]=A[p2]; A[p2]=2; while(A[p2]==2)p2--; } } else p++; } } };
LeetCode: Sort Colors [075]