75. Sort Colors (M)

Sort Colors (M)

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?

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题意

将一个只包含0、1、2的数组排序，不允许使用内置的排序函数。

思路

经典的“荷兰旗??”问题。

最简单的做法是先遍历一遍统计0、1、2的个数，再遍历一遍逐个按个数输出0、1、2。

只遍历一遍的方法，实际上是对最基础的快速排序方法进行重复元素优化：将数组分为三个区间，左区间存放比目标值小的元素，右区间存放比目标值大的元素，中部区间存放与目标值相同的元素，每次递归只递归左区间和右区间。

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代码实现 - 快排优化

class Solution {
    public void sortColors(int[] nums) {
        sort(nums, 0, nums.length - 1);
    }

    private void sort(int[] nums, int start, int end) {
        if (start >= end) {
            return;
        }

        int i = start;						// 当前遍历元素
        int left = start, right = end;		// left左侧元素小于目标值，right右侧元素大于目标值
        int temp = nums[start];				// 目标值

        while (i <= right) {
            if (nums[i] < temp) {
                swap(nums, i++, left++);	// 当前元素小于目标值，移到左侧区间
            } else if (nums[i] > temp) {
                swap(nums, i, right--);		// 当前元素大于目标值，移到右侧区间
            } else {
                i++;						// 当前元素等于目标值，保留在中间区间中不移动
            }
        }

        sort(nums, start, left - 1);
        sort(nums, right + 1, end);
    }


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

代码实现 - 统计排序

class Solution {
    public void sortColors(int[] nums) {
        // 统计个数
        int[] count = new int[3];
        for (int i = 0; i < nums.length; i++) {
            count[nums[i]]++;
        }
        // 重新填充数组
        int i = 0;
        for (int j = 0; j < 3; j++) {
            while (count[j]-- != 0) {
                nums[i++] = j;
            }
        }
    }
}