【Leetcode】Sort Colors

最新推荐文章于 2025-09-09 20:07:09 发布

转载最新推荐文章于 2025-09-09 20:07:09 发布 · 111 阅读

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本文提供了一种解决荷兰国旗问题的Java算法，该算法在一次遍历中完成排序，并仅使用常数空间。重点讨论了如何在不使用排序库函数的情况下，通过计数不同颜色的元素数量并覆盖数组来解决问题。

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?

0,1,2分别代表3种颜色，同一种颜色的代码相邻，也叫荷兰国旗问题虽然简单但也要想清楚 swap的问题。

Java:

1. http://www.cnblogs.com/feiling/p/3256118.html

public class Solution {
    public void sortColors(int[] A) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int len = A.length;
        int i = 0, redIdx = 0, blueIdx = len - 1;
        while(i < blueIdx + 1){
            if(A[i] == 0){
                swap(A, i, redIdx);
                redIdx ++;
                i ++;
            } else if(A[i] == 2){
                swap(A, i, blueIdx);
                blueIdx --;
            } else{
                i++;
            }
        }
    }
    
    public void swap(int[] A, int i, int idx){
        int tmp = A[i];
        A[i] = A[idx];
        A[idx] = tmp;
    }
}

2. hash http://blog.youkuaiyun.com/xiaozhuaixifu/article/details/12888435

public class Solution {
    public void sortColors(int[] A) {
        int[] hash= new int[3];
        if(A.length==0)
        return;
        for(int key:A)
        {
            hash[key]++;
        }
        int len=0;
        for(int i=0;i<3;i++)
        {
            int cnt=hash[i];
            if(cnt>0)
            {
                for(int j=0;j<cnt;j++)
                {
                    
                    A[len++]=i;
                    
                }
            }
        }
    }
}