75. Sort Colors

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:

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?

Problem URL

---

Solution

0，1，2三种颜色，将相同颜色排列在一起。

Using an array to store count, and use another for loop to assign value to corresponding index. Remember count[0] + count[1] is right.

Code

class Solution {
    public void sortColors(int[] nums) {
        int[] count = new int[3];
        for (int i = 0; i < nums.length; i++){
            count[nums[i]]++;
        }
        
        for (int i = 0; i < nums.length; i++){
            if (i < count[0]){
                nums[i] = 0;
            }
            else if (i < count[0] + count[1]){
                nums[i] = 1;
            }
            else{
                nums[i] = 2;
            }
        }
    }
}

Time Complexity: O(n)
Space Complexity: O(1)

---

Review

Another approach is one pass solution.There are two things need to be concerned. It is all about right pointer. First, I <= right, for the reason that, we definitely pass left, but right is unsorted, so we need I = right. Second is that when we swap 2 with right, the value of right is not sure, so we need I-- to perform another swap.

class Solution {
    public void sortColors(int[] nums) {
        int left = 0, right = nums.length - 1, i = 0;
        while (i <= right){
            if (nums[i] == 0){
                nums[i] = nums[left];
                nums[left] = 0;
                left++;
            }
            if (nums[i] == 2){
                nums[i] = nums[right];
                nums[right] = 2;
                right--;
                i--;
            }
            i++;
        }
    }
}