[Leetcode]sort Color

[timu]                                                                                    <p>Given an array with <em>n</em> 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.</p><p>Here, we will use the integers 0, 1, and 2 to represent the color red, white, and blue respectively.</p><p><strong>Note:</strong>You are not suppose to use the library's sort function for this problem.</p><p class="showspoilers"></p><p><strong>Follow up:</strong>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.</p>Could you come up with an one-pass algorithm using only constant space?[silu]two pass: first pass: count the number of 0,1,2second pass: fill the array with number of 0,1,2one pass solution:keep two pointers to point the beginning of the 0,and 2;public void sortColors(int[] nums) {    // 1-pass    int p1 = 0, p2 = nums.length - 1, index = 0;    while (index <= p2) {        if (nums[index] == 0) {            nums[index] = nums[p1];            nums[p1] = 0;            p1++;        }        if (nums[index] == 2) {            nums[index] = nums[p2];            nums[p2] = 2;            p2--;            <span style="background-color: rgb(255, 204, 204);">index--;</span>        }        index++;    }}public void sortColors(int[] nums) {    // 2-pass    int count0 = 0, count1 = 0, count2 = 0;    for (int i = 0; i < nums.length; i++) {        if (nums[i] == 0) {count0++;}        if (nums[i] == 1) {count1++;}        if (nums[i] == 2) {count2++;}    }    for(int i = 0; i < nums.length; i++) {        if (i < count0) {nums[i] = 0;}        else if (i < count0 + count1) {nums[i] = 1;}        else {nums[i] = 2;}    }}