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 supposed 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?
class Solution {
public:
void sortColors(vector<int>& nums)
{
// int a[3]={0};
// for(auto i:nums)
// {
// a[i]++;
// }
// int j=0;
// for(int k=0;k<a[0];++k)
// nums[j++]=0;
// for(int k=0;k<a[1];++k)
// nums[j++]=1;
// for(int k=0;k<a[2];++k)
// nums[j++]=2;
int n = A.size();
int n0 = -1, n1 = -1, n2 = -1;
for (int i = 0; i < n; ++i) {
if (A[i] == 0)
{
A[++n2] = 2; A[++n1] = 1; A[++n0] = 0;
}
else if (A[i] == 1)
{
A[++n2] = 2; A[++n1] = 1;
}
else if (A[i] == 2)
{
A[++n2] = 2;
}
}
};
class Solution {
public:
void sortColors(vector<int>& nums)
{
// int a[3]={0};
// for(auto i:nums)
// {
// a[i]++;
// }
// int j=0;
// for(int k=0;k<a[0];++k)
// nums[j++]=0;
// for(int k=0;k<a[1];++k)
// nums[j++]=1;
// for(int k=0;k<a[2];++k)
// nums[j++]=2;
}
};
