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

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?

题目要求是，时间复杂度O(n) 空间复杂度O(1) 对于没有听过三路快排的窝来说，这个medium比hard难得多好趴。。。

先说原理，

快速排序就是找一个基准v，通过双向指针，把<v的值放在左边，>=v的值放在右边，然后递归。

三路，就是 <v， =v， >v。那么如何分成三分呢？

设有一个数组 [0...n] 遍历数组，index 表示遍历到的位置。

找两个分界位置，lt 和 rt 用来表示 =v 和 >v 的最小位置。

如果遇到小于 v 的就让其和lt交换，同时将 lt++。等于 v 不需要做操作。大于 v 就将 lt-- 其和 lt 作交换

之后将 <v 和 >v 的两段递归排序就可以啦。

那么这题，只是三路排序的一个思路~~

/*
 * @lc app=leetcode id=75 lang=javascript
 *
 * [75] Sort Colors
 */
/**
 * @param {number[]} nums
 * @return {void} Do not return anything, modify nums in-place instead.
 */
var sortColors = function(nums) {
    let n = nums.length;
    let lt = 0, // =v 的第一个
        rt = n; // >v 的第一个

    let index = 0;
    let v = 1;

    while (index < n && index < rt) {
        if (nums[index] > v) {
            rt--;
            swap(index, rt);
        } else if (nums[index] === v) {
            index++;
        } else if (nums[index] < v) {
            swap(lt, index);
            lt++;
            index++;
        }
    }

    function swap(i, j) {
        let t = nums[i];
        nums[i] = nums[j];
        nums[j] = t;
    }
};