Leetcode：Sort Colors

最新推荐文章于 2018-09-19 08:24:21 发布

卧龙_byr

最新推荐文章于 2018-09-19 08:24:21 发布

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文章标签： Sort Colors

本文链接：https://blog.youkuaiyun.com/wolongdede/article/details/40211203

本文探讨了一种解决特定数组排序问题的方法，通过多思考，我们能够找到多种解决方案。重点介绍了利用计数排序实现一次遍历的算法，并提供了一个实际的Java代码实现。讨论了该方法的效率和适用场景，特别强调了它在处理特定类型数组排序任务时的优势。

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做了这题的感想就是：多思考，其实办法会有很多的。

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?

比较正式的解法：

public class Solution {
    public void sortColors(int[] a) {
        if(a == null || a.length <= 1)
            return;
        
        int pl = 0;
        int pr = a.length - 1;
        int i = 0;
        while(i <= pr){
            if(a[i] == 0){
                swap(a, pl, i);
                pl++;
                i++;
            }else if(a[i] == 1){
                i++;
            }else{
                swap(a, pr, i);
                pr--;
            }
        }
    }
    
    private void swap(int[] a, int i, int j){
        int tmp = a[i];
        a[i] = a[j];
        a[j] = tmp;
    }
}

在给出网友的一种解法，感觉有点“旁门左道”，但是能很好的解决问题，确实是个好方法。

  public void sortColors(int[] A) {


    int i=-1, j=-1, k=-1;

    for(int p = 0; p < A.length; p++)
    {
        if(A[p] == 0)
        {
            A[++k]=2;
            A[++j]=1;
            A[++i]=0;
        }
        else if (A[p] == 1)
        {
            A[++k]=2;
            A[++j]=1;

        }
        else if (A[p] == 2)
        {
            A[++k]=2;
        }
    }

}