leetcode -- 75. Sort Colors

最新推荐文章于 2025-09-05 23:16:08 发布

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最新推荐文章于 2025-09-05 23:16:08 发布

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本文详细解析了如何在不使用库排序函数的情况下，对由红、白、蓝三种颜色（用0、1、2表示）组成的数组进行原地排序，以达到相同颜色相邻且按红白蓝顺序排列的目标。文章提供了两种解决方案，包括计数排序的两遍法及一过法，并附带了具体的代码实现。

题目描述

题目难度：Medium
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?

AC代码

方法1：分别记录0、1、2这三个数字出现的次数，然后按照顺序写进数组；
方法2：先把0、1看作一类，2看作另一类，总共两类，遍历数组，让2全部都处于数组的右端；然后剩下的数组里只有0、1两类，按照上述步骤即可完成。三类不好分，但是两类很好分。这种思维值得学习。

class Solution {
    public void sortColors(int[] nums) {
        if(nums == null || nums.length <= 1) return;
        swapZeroAndTwo(nums);
        int i = nums.length - 1;
        while(i >= 0 && nums[i] == 2) i--;
        if(i <= 0) return;
        swapZeroAndOne(nums, i);
    }
    
    private void swapZeroAndTwo(int[] nums){
        int low = 0, high = nums.length - 1;
        while(low < high){
            if(nums[low] != 2) low++;
            else if(nums[high] == 2) high--;
            else
            {
                int tmp = nums[low];
                nums[low] = nums[high];
                nums[high] = tmp;
            }
        }
    }
    
    private void swapZeroAndOne(int[] nums, int end){
        int low = 0, high = end;
        while(low < high){
            if(nums[low] != 1) low++;
            else if(nums[high] == 1) high--;
            else
            {
                int tmp = nums[low];
                nums[low] = nums[high];
                nums[high] = tmp;
            }
        }
    }
}