Leetcode-75.Sort Colors

地址：https://leetcode.com/problems/sort-colors/

描述：

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.

實例：

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?

方法1：

 // 時間複雜度O(n)
    // 空間複雜度 O(k)
    public void sortColors(int[] nums) {
        int[] count = new int[3];

        for (int i = 0; i < nums.length; i++) {
            if (nums[i] >= 0 && nums[i] <= 2) {
                count[nums[i]]++;
            }
        }

        int index = 0;
        for (int i = 0; i < count[0]; i++) {
            nums[index++] = 0;
        }

        for (int i = 0; i < count[1]; i++) {
            nums[index++] = 1;
        }

        for (int i = 0; i < count[2]; i++) {
            nums[index++] = 2;
        }
    }

方法2：

    // 時間複雜度O(n)
    // 空間複雜度 O(1)
    public void sortColors_2(int[] nums) {
        int zero = -1;  //nums[0,zero]是0
        int two = nums.length; // nums[two...n-1]是2

        for (int i = 0; i < two; i++) {
            if (nums[i] == 1) {
                i++;
            } else if (nums[i] == 2) {
                swap(nums, i, --two);
            } else {
                if (nums[i] == 0) {
                    swap(nums, ++zero, i++);
                }
            }
        }
    }