LeetCode

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.

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?

思路：

保持两个指针，一个指向最前端，一个指向最后端，然后扫描，遇到0,2就交换元素

package sort;

public class SortColors {

    public void sortColors(int[] nums) {
        int n;
        if (nums == null || (n = nums.length) < 2) return;
        int zeroIndex = 0;
        int twoIndex = n - 1;
        for (int i = 0; i <= twoIndex; ++i) {
            if (nums[i] == 0) {
                swap(nums, i, zeroIndex);
                ++zeroIndex;
            } else if (nums[i] == 2) {
                swap(nums, i, twoIndex);
                --twoIndex;
                --i;
            }
        }
    }
    
    private void swap(int[] A, int i, int j) {
        int tmp = A[i];
        A[i] = A[j];
        A[j] = tmp;
    }
    
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        int[] A = { 2, 1, 0, 1, 2, 0, 0, 2, 1, 1, 2, 0 };
        SortColors s = new SortColors();
        s.sortColors(A);
        for (int i : A)
            System.out.println(i);
    }

}