LeetCode | Sort Colors

題目

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?

三色旗問題，題目要求了不用計數排序，那就用正統的類似三路快排的方法（解法1）。同時leetcode的討論區給出了一種很“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;
	}
}

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;
			}
		}
	}
}