LeetCode – Sort Colors (Java)

 

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.

Java Solution 1 - Counting Sort

Check out this animation to understand how counting sort works.

public void sortColors(int[] nums) {
    if(nums==null||nums.length<2){
        return;
    }
 
    int[] countArray = new int[3];
    for(int i=0; i<nums.length; i++){
        countArray[nums[i]]++;
    }
 
    for(int i=1; i<=2; i++){
        countArray[i]=countArray[i-1]+countArray[i];
    }
 
    int[] sorted = new int[nums.length];
    for(int i=0;i<nums.length; i++){
        int index = countArray[nums[i]]-1;
        countArray[nums[i]] = countArray[nums[i]]-1;
        sorted[index]=nums[i];
    }
 
    System.arraycopy(sorted, 0, nums, 0, nums.length);
}

Java Solution 2 - Improved Counting Sort

In solution 1, two arrays are created. One is for counting, and the other is for storing the sorted array (space is O(n)). We can improve the solution so that it only uses constant space. Since we already get the count of each element, we can directly project them to the original array, instead of creating a new one.

public void sortColors(int[] nums) {
    if(nums==null||nums.length<2){
        return;
    }
 
    int[] countArray = new int[3];
    for(int i=0; i<nums.length; i++){
        countArray[nums[i]]++;
    }
 
    int j = 0;
    int k = 0;
    while(j<=2){
        if(countArray[j]!=0){
            nums[k++]=j;
            countArray[j] = countArray[j]-1;
        }else{
            j++;
        }
    }
}

Related posts:

  1. LeetCode – Summary Ranges (Java)
  2. LeetCode – Next Permutation (Java)
  3. LeetCode – Search for a Range (Java)
  4. LeetCode – 3Sum
Category >> Algorithms >> Interview  
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  1. up23 on 2016-1-2

    Another take at solution 2. Sorry it’s in PHP.


    function sortColors(&$inputs) {
    if (!$inputs || count($inputs) $v) {
    for ($j = 0; $j < $v; $j ++) {
    $inputs[$i ++] = $k;
    }
    }
    }

  2. Vasyl Grygoryev on 2016-1-3

    More simple solution for 1st case (two-pass algorithm):

    public void sortColors(int[] nums) {
    if (nums == null || nums.length < 2)
    return;

    int[] color = new int[3];

    for (int num : nums)
    color[num]++;

    int i = 0;
    for (int j = 0; j 0)
    nums[i++] = j;
    }

    One-pass algorithm:

    public void sortColors(int[] nums) {
    if (nums == null || nums.length < 2)
    return;

    int red = 0;
    int blue = nums.length - 1;
    int i = 0;

    while (i < blue + 1)
    if (nums[i] == 0) {
    nums[i] = nums[red];
    nums[red++] = 0;
    i++;
    }
    else if (nums[i] == 2) {
    nums[i] = nums[blue];
    nums[blue--] = 2;
    }
    else i++;
    }

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