75. Sort Colors

Problem:

Given an array with n objects colored red, white or blue, sort them in-placeso 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.

Note: You are not suppose to use the library’s sort function for this problem.

Example:

Input: [2,0,2,1,1,0]
Output: [0,0,1,1,2,2]

Follow up:

• 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.
• Could you come up with a one-pass algorithm using only constant space?

Solution:

One-pass algorithm.

Take the idea of the partition algorithm from quick sort. Use 1 as pivot.

Count the number of sorted 0s and 2s so that we know where to swap.

/**
 * @param {number[]} nums
 * @return {void} Do not return anything, modify nums in-place instead.
 */
let sortColors = function (nums) {
  const len = nums.length;
  let zeroEnd = 0;
  let twoStart = len - 1;
  let i = 0;
  while (i <= twoStart) {
    if (nums[i] === 0 && i !== zeroEnd) {
      const t = nums[i];
      nums[i] = nums[zeroEnd];
      nums[zeroEnd] = t;
      zeroEnd++;
    } else if (nums[i] === 2 && i !== twoStart) {
      const t = nums[i];
      nums[i] = nums[twoStart];
      nums[twoStart] = t;
      twoStart--;
    } else {
      i++;
    }
  }
};

☆: .｡. o(≧▽≦)o .｡.:☆☆: .｡. o(≧▽≦)o .｡.:☆☆: .｡. o(≧▽≦)o .｡.:☆

☆: .｡. o(≧▽≦)o .｡.:☆☆: .｡. o(≧▽≦)o .｡.:☆