1. Leetcode 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]

2. 分析思路1

通过第一次遍历记录0、1、2对应的个数（桶排序），然后第二次遍历将0、1、2进行填入即可。

2.1 代码实现

2.1.1 C++版本

class Solution {
public:
    void sortColors(vector<int>& nums) {
        int a[3] = {0}; //统计0 1 2出现的次数
        for (int i=0; i<nums.size(); ++i) {
            ++a[nums[i]];
        }

        int base = 0;
        int digit = 0;

        #外层循环表示的是几个数
        for (int i=0; i<3; ++i) {
            int offset = 0;

            for (int j=0; j<a[i]; ++j) {
                nums[base + offset] = digit;
                ++offset;
            }

            base += offset;
            ++digit;
        }

    }
};

2.1.2 Java版本

class Solution {
    public void sortColors(int[] nums) {
        int[] a = new int[3];
        for(int i=0; i<nums.length; ++i) {
            ++a[nums[i]];
        }

        int base = 0;
        int digit = 0;

        for(int i=0; i<3; ++i) {
            int offset = 0;
            for (int j=0; j<a[i]; ++j) {
                nums[base + offset] = digit;
                ++offset;
            }

            base += offset;
            ++digit;
        }
    }
}

2.1.3 Python2版本

class Solution(object):
    def sortColors(self, nums):
        num_counts = [0, 0, 0]

        for i in nums:
            num_counts[i] += 1

        base = 0
        digit = 0

        for i in range(0, 3):
            offset = 0
            for j in range(0, num_counts[i]):
                nums[base + offset] = digit
                offset += 1

            base += offset
            digit += 1

3. 分析思路2

  参考链接为https://www.geeksforgeeks.org/sort-an-array-of-0s-1s-and-2s/

思路类似于三路快排，把数组arr[0, N-1] 划分为四个区间，如下所示：

  1. a[0, i-1] zeroes
  2. a[i, k-1] ones
  3. a[k, j] unknown
  4. a[j+1, len(a)-1] twos

  初始值时，i=0，j=len(a)-1，而k=0，所以zeros、ones、twos三个区间均为空值。最终要使得a[k, j]区间为空，而k > j 时区间为空，所以循环的满足条件为k <= j。

3.1 代码实现

class Solution(object):
    def sortColors(self, nums):
        """
        :type nums: List[int]
        :rtype: void Do not return anything, modify nums in-place instead.
        """
        low = 0
        mid = 0
        high = len(nums) - 1

        # low <= mid <= high
        while mid <= high:
            if nums[mid] == 0:
                nums[low], nums[mid] = nums[mid], nums[low]
                low += 1
                mid += 1

            elif nums[mid] == 1:
                mid += 1

            else:
                nums[high], nums[mid] = nums[mid], nums[high]
                high -= 1
class Solution {
public:
    void sortColors(vector<int>& nums) {
        int i = -1;
        int j = 0;
        int k = nums.size();
        
        while (j < k){
            if (nums[j] == 1) {
                ++j;
            } else if (nums[j] == 0) {
                swap(nums[++i], nums[j++]);
            } else{
                swap(nums[j], nums[--k]);
            }
        }
            
    }
};