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Given an array nums
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.
We will use the integers 0
, 1
, and 2
to represent the color red, white, and blue, respectively.
You must solve this problem without using the library’s sort function.
Example 1:
Input: nums = [2,0,2,1,1,0]
Output: [0,0,1,1,2,2]
Example 2:
Input: nums = [2,0,1]
Output: [0,1,2]
Constraints:
n == nums.length
1 <= n <= 300
nums[i]
is either 0
, 1
, or 2
.Follow up: Could you come up with a one-pass algorithm using only constant extra space?
def sortColors(self, nums):
red, white, blue = 0, 0, len(nums)-1
while white <= blue:
if nums[white] == 0:
nums[red], nums[white] = nums[white], nums[red]
white += 1
red += 1
elif nums[white] == 1:
white += 1
else:
nums[white], nums[blue] = nums[blue], nums[white]
blue -= 1
class Solution {
public:
void sortColors(vector<int>& nums)
{
int tmp = 0, low = 0, mid = 0, high = nums.size() - 1;
while(mid <= high)
{
if(nums[mid] == 0)
{
tmp = nums[low];
nums[low] = nums[mid];
nums[mid] = tmp;
low++;
mid++;
}
else if(nums[mid] == 1)
{
mid++;
}
else if(nums[mid] == 2)
{
tmp = nums[high];
nums[high] = nums[mid];
nums[mid] = tmp;
high--;
}
}
}
};
public void sortColors(int[] nums) {
// 1-pass
int p1 = 0, p2 = nums.length - 1, index = 0;
while (index <= p2) {
if (nums[index] == 0) {
nums[index] = nums[p1];
nums[p1] = 0;
p1++;
}
if (nums[index] == 2) {
nums[index] = nums[p2];
nums[p2] = 2;
p2--;
index--;
}
index++;
}
}
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