Sort Colors

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.

click to show follow up.

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?
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题目中说了一种方法，数出每种颜色的个数，然后直接覆盖就行，只是需要两次循环。要在一次循环中完成，我们设置两个边界，l和r，遇到0就和l值交换，遇到2就和r值交换，遇到1则不进行任何操作。要注意的是r–的同时要i–，因为交换过来的值可能出现0，因此要重新检测。代码如下：

class Solution {
public:
void sortColors(vector<int>& nums) {
int l=0, r=nums.size()-1;
for(int i=l;i<=r;i++){
if(nums[i]==0){
int tmp=nums[i];
nums[i]=nums[l];
nums[l]=tmp;
l++;
}
if(nums[i]==2){
int tmp=nums[i];
nums[i]=nums[r];
nums[r]=tmp;
r--;
i--;
}
}
}
};

还有一种平移插入的方法。代码如下：

public void sortColors(int[] A) {

int i=-1, j=-1, k=-1;

for(int p = 0; p < A.length; p++)
{
if(A[p] == 0)
{
A[++k]=2;
A[++j]=1;
A[++i]=0;
}
else if (A[p] == 1)
{
A[++k]=2;
A[++j]=1;

}
else if (A[p] == 2)
{
A[++k]=2;
}
}

}