LeetCode - Sort Colors
Problem description
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 suppose to use the library’s sort function for this problem.
Example:
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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?
Analysis
The general idea is to insert 0 and 2 into correct place.
Lets have a pointer for 0, and if the current is 0, we can swap current item with zero and update zero and current. if the current is 2, we can swap and update two.
we should not update current if we have 2. Because current is always equal to or bigger than zero, so the swapped item would be 0 or 1 and it’s ok. But current is always less or equal to two, so we may swap a 0 with a 2. And we need run the check again.
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class Solution {
public void sortColors(int[] nums) {
int zero = 0;
int two = nums.length - 1;
int i = 0;
while (zero <= i && i <= two){
if (nums[i] == 0){
// if (swap(nums, i, zero)){
// i--;
// }
swap(nums, i, zero);
zero++;
}
else if (nums[i] == 2){
// if (swap(nums, i, two))
// i--;
swap(nums, i, two);
two--;
i--;
}
i++;
}
}
boolean swap(int[] nums, int i, int j){
if (i == j){
return false;
}
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
return true;
}
}