75 Sort Colors

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?

由于0, 1, 2 非常紧凑，首先想到计数排序(counting sort)，但需要扫描两遍，不符合题目要求。

由于只有三种颜色，可以设置两个index，一个是red的index，一个是blue的index，两边往中间走。时间复杂度 O(n) ，空间复杂度 O(1) 。

第3种思路，利用快速排序里 partition 的思想，第一次将数组按0分割，第二次按1 分割，排序完毕，可以推广到 n 种颜色，每种颜色有重复元素的情况。

class Solution {
    public void sortColors(int[] nums) {
        int red = 0, blue = nums.length - 1;
        for (int i = 0; i < blue + 1;) {
            if (nums[i] == 0) {
                swap(nums, i++, red++);
            } else if (nums[i] == 2) {
                swap(nums, i, blue--);
            } else {
                i++;
            }
        }
    }
    private void swap(int[] nums, int i, int j) {
        int tmp = nums[i];
        nums[i] = nums[j];
        nums[j] = tmp;
    }
}

/**
 *
 * @author dongb
 * Sort Colors
 * Rebuild partition()
 * TC O(n). SC O(1)
 * 
 */
public class Solution {
    public void sortColors(int[] nums) {
        partition(nums, partition(nums, 0, nums.length, new EqualTo(0)), nums.length,
                new EqualTo(1));
    }
    
    private static int partition(int[] nums, int begin, int end, EqualTo predicate) {
        int pos = begin;
        
        for (; begin != end; begin++) {
            if (predicate.apply(nums[begin]))
                swap(nums, begin, pos++);
        }
        return pos;
    }
    
    static class EqualTo {
        private final int target;
        public EqualTo(int target) {
            this.target = target;
        }
        public boolean apply(int x) {
            return x == target;
        }
    }
    
    private static void swap(int[] nums, int i, int j) {
        if (i == j) return;
        int tmp = nums[i];
        nums[i] = nums[j];
        nums[j] = tmp;
    }
}