[leetcode] Sort Colors

Given an array with n 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.

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.

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 an one-pass algorithm using only constant space?

https://oj.leetcode.com/problems/sort-colors/

思路1： counting sort. 如题目中所示，需要2 passes。

思路2：双指针法。（其实是algorithms里讲的的quicksort中的three-way-partition部分，详见这里）

• 　　两指针一前一后分别指向1的开始元素和1的结束元素。

思路2代码

import java.util.Arrays;

public class Solution {
    public void sortColors(int[] A) {
        if (A == null || A.length <= 1)
            return;
        int n = A.length;
        int lt = 0;
        int gt = n - 1;
        int i = 0;

        while (i <= gt) {
            if (A[i] == 1) {
                i++;
            } else if (A[i] == 0) {
                swap(A, lt++, i++);
            } else {
                swap(A, gt--, i);
            }

        }

    }

    private void swap(int[] a, int i, int j) {
        int tmp = a[i];
        a[i] = a[j];
        a[j] = tmp;
    }

    public static void main(String[] args) {
        int[] A = { 1, 0, 2, 0, 1 };
        new Solution().sortColors(A);
        System.out.println(Arrays.toString(A));
    }
}

 第二遍记录：

　　注意循环的条件 （i<=gt）等号的边界情况要注意。

　　注意lt,gt代表的含义　　　　

第四遍记录：

left指针左边是排好的0，right右边是排好的2；

注意遍历指针的终止条件，i<=right，等于不要忘记了。

public class Solution {
    public void sortColors(int[] a) {
        int n =a.length;
        int left=0,right=n-1;
        int i=0;
        while(i<=right){
            if(a[i]==0){
                swap(a,left++,i++);
            }else if(a[i]==2){
                swap(a,i,right--);
            }else{
                i++;
            }
            
        }
    }
    
    private void swap(int[]a,int i,int j){
        if(i!=j){
            int tmp = a[i];
            a[i]=a[j];
            a[j]=tmp;
        }
    }
    
    
}