75. Sort Colors

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?

要求我们只用一次遍历就得出结果。参考了来offer的教程 https://www.youtube.com/watch?v=yTwW8WiGrKw&list=PLTNkreZiUTIL-S_VJBLRxlmGktAQtla-m&index=3

采用了三个挡板来分割四种类型的字母（数字）——0，1，xxx（未确定），2

　类似于　000 | 11| xxx | 22

　　　　0——[0,i)

　　　　1——[i,j)

　　　　x——[j,k]

　　　　2——(k,length-1]

然后用设index i，j，k分别代表0的末位，未知数字的首位和2的首位。初始值为i=0, j=0, k=length-1;

这样，用一次遍历，检查 nums[ j ]的值，是0就和 nums[ i ]交换位置，同时 i++，j++。

是1就 j++。

是2就和nums[ k ]交换位置，同时k--。

代码如下

class Solution {
    public void sortColors(int[] nums) {
        if(nums.length <=1 || nums == null) return;
        int i = 0, j = 0, k = nums.length -1;
        while(j<=k){
            if(nums[j]==0){
                swap(i++,j++,nums);
            }
            else if(nums[j] == 1){
               j++; 
            }
            else if(nums[j] == 2){
                swap(j,k--,nums);
            }
        }
    }
    public void swap(int i, int j, int nums[]){
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
    }
}

 https://leetcode.com/problems/sort-colors/discuss/26472/Share-my-at-most-two-pass-constant-space-10-line-solution

low-high two pointers.