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?
分析:三個i、j、k 標示量,i 表示其前面的都是0, j到k之間的爲1, k右邊的都爲2
class Solution {
public:
void sortColors(int A[], int n) {
int i = 0; // 0 pointer
int j = n - 1; // 1 pointer
int k = n - 1; // 2 pointer
while (i <= j)
{
if (A[i] == 2)
{
swap(A[i], A[k]);
k--;
if (k < j) j = k;
}
else if (A[i] == 1)
{
swap(A[i], A[j]);
j--;
}
else if (A[i] == 0) {
i++;
}
}
}
};