LeetCode | Sort Colors

最新推荐文章于 2025-08-11 14:43:21 发布

Nero_pf

最新推荐文章于 2025-08-11 14:43:21 发布

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分类专栏： LeetCode 文章标签： LeetCode java 三色旗

本文链接：https://blog.youkuaiyun.com/perfect8886/article/details/20835895

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本文探讨了解决三色旗问题的两种方法：一种基于类似快速排序的算法，另一种则是更为直观直接的方法。这两种算法均不使用额外的排序函数，并且在常数空间复杂度内完成任务。

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题目

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?

分析

三色旗问题，题目要求了不用计数排序，那就用正统的类似三路快排的方法（解法1）。同时leetcode的讨论区给出了一种很“直接”的方法（解法2）。

解法1

public class SortColors {
	public void sortColors(int[] A) {
		if (A == null || A.length == 0) {
			return;
		}

		int red = -1;
		int blue = A.length;
		int p = 0;
		while (p < blue) {
			if (A[p] == 0) {
				swap(A, ++red, ++p);
			} else if (A[p] == 2) {
				swap(A, --blue, p);
			} else {
				++p;
			}
		}
	}

	private final void swap(int[] A, int i, int j) {
		int temp = A[i];
		A[i] = A[j];
		A[j] = temp;
	}
}

解法2

public class SortColors {
	public void sortColors(int[] A) {
		if (A == null || A.length == 0) {
			return;
		}

		int red = -1;
		int white = -1;
		int blue = -1;
		for (int i = 0; i < A.length; ++i) {
			if (A[i] == 0) {
				A[++blue] = 2;
				A[++white] = 1;
				A[++red] = 0;
			} else if (A[i] == 1) {
				A[++blue] = 2;
				A[++white] = 1;
			} else {
				A[++blue] = 2;
			}
		}
	}
}