# 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 - Counting Sort

We can get the count of each element and project them to the original array.

```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++; } } }```
Category >> Algorithms >> Interview
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• alexwest11

obv it can be done w/out add memory or counting

• Vasyl Grygoryev

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++; } ```

• up23

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; } } } ```