CODE 58: Sort Colors

最新推荐文章于 2025-06-15 21:19:16 发布

原创最新推荐文章于 2025-06-15 21:19:16 发布 · 653 阅读

0 ·

CC 4.0 BY-SA版权

leetcode 专栏收录该内容

154 篇文章

订阅专栏

本文介绍了一种不使用内置排序函数的颜色排序算法。该算法的目标是对由红、白、蓝三种颜色组成的数组进行排序，使得相同颜色的元素相邻，并按红、白、蓝的顺序排列。通过计数排序的方式，在一次遍历中统计每种颜色的数量，然后重新写入数组来实现排序。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

click to show follow up.

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?

WORST CODE:

	public void sortColors(int[] A) {
		// Note: The Solution object is instantiated only once and is reused by
		// each test case.
		// 012012012012
		int start = 0;
		int end = A.length - 1;
		while (start < A.length && A[start] == 0) {
			start++;
		}
		while (end > 0 && A[end] == 2) {
			end--;
		}
		for (int mid = start; mid <= end; mid++) {
			if (A[mid] == 0) {
				int tmp = A[mid];
				A[mid] = A[start];
				A[start] = tmp;
				start++;
				while (start < A.length && A[start] == 0) {
					start++;
				}
				mid = start - 1;
			} else if (mid != end && A[mid] == 2) {
				int tmp = A[mid];
				A[mid] = A[end];
				A[end] = tmp;
				end--;
				mid--;
				while (end > 0 && A[end] == 2) {
					end--;
				}
			}
		}
	}