One pointer. Two pass. Just be careful:

  1. for loop: i is on color genre, not nums.length;
  2. self-increment each time: count++ is independent of i, j.
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    public void sortColors(int[] nums) {
            int[] colors = new int[3];
            for (int num : nums) colors[num]++;
            for (int i = 0, count = 0; i < 3; ++i) {
                for (int j = 0; j < colors[i]; ++j) {
                    nums[count++] = i;
                }
            }
        }

Two pointers. Iterate through; swap red pointer number with current 0, swap blue pointer number with current 2.

attention: for ... i <= blue

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class Solution {
    public void sortColors(int[] nums) {
        int red = 0, blue = nums.length - 1;
        for (int i = 0; i <= blue; ++i) {
            if (nums[i] == 0) {
                int tmp = nums[red];
                nums[red] = nums[i];
                nums[i] = tmp;
                red++;
            }
            else if (nums[i] == 2) {
                int tmp = nums[blue];
                nums[blue] = nums[i];
                nums[i] = tmp;
                i--;
                blue--;
            }
            
        }
    }
}
  1. blue:2, 交换完成之后i--, 因为要继续检查被2交换回来的数;
  2. red:0, 没有i++, 不用检查和0交换的数，因为肯定是1;
  3. 上面的，为什么“和0交换的一定是1?” 因为如果是2，直接进去后面一个if了，进去的结果是2会被都移动到最后面，所以从0左边开始碰到的肯定都是1，而1不用做任何处理