Package g0001_0100.s0075_sort_colors

Class Solution

java.lang.Object
g0001_0100.s0075_sort_colors.Solution
public class Solution extends Object
75 - Sort Colors.

Medium

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]

Example 3:

Input: nums = [0]

Output: [0]

Example 4:

Input: nums = [1]

Output: [1]

Constraints:

  • n == nums.length
  • 1 <= n <= 300
  • nums[i] is 0, 1, or 2.

Follow up: Could you come up with a one-pass algorithm using only constant extra space?

To solve the “Sort Colors” problem in Java with the Solution class, follow these steps:

  1. Define a method sortColors in the Solution class that takes an array of integers nums as input and sorts it in-place according to the colors red, white, and blue.
  2. Initialize three pointers: low, mid, and high. low points to the beginning of the array, mid points to the current element being processed, and high points to the end of the array.
  3. Loop while mid is less than or equal to high:
    • If nums[mid] is 0, swap nums[low] with nums[mid], increment low and mid.
    • If nums[mid] is 1, increment mid.
    • If nums[mid] is 2, swap nums[mid] with nums[high], decrement high.
  4. After the loop, the array will be sorted in-place according to the colors red, white, and blue.

Here’s the implementation of the sortColors method in Java:

class Solution {
    public void sortColors(int[] nums) {
        int low = 0;
        int mid = 0;
        int high = nums.length - 1;
        
        while (mid <= high) {
            if (nums[mid] == 0) {
                swap(nums, low, mid);
                low++;
                mid++;
            } else if (nums[mid] == 1) {
                mid++;
            } else {
                swap(nums, mid, high);
                high--;
            }
        }
    }
    
    private void swap(int[] nums, int i, int j) {
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
    }
}

This implementation sorts the array in-place using a one-pass algorithm with constant extra space. It iterates through the array and swaps elements as needed to group them according to their colors. The time complexity of this solution is O(n), where n is the length of the array.

  • Constructor Details

    • Solution

      public Solution()

  • Method Details

    • sortColors

      public void sortColors(int[] nums)