75. Sort Colors

題目：
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.

click to show follow up.

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?

解題：
能夠參考EPI的14.8, 這題比較簡單，就沒有用書裏的解法，follow up的思想就是交換，既然只能one pass，那就一次至少搞定一個數啦
解法1:

public void Color(int[] nums, int color, int start, int len) {
        for (int i = start; i < start + len; i++) {
            nums[i] = color;
        }
    }
    public void sortColors(int[] nums) {
        if (nums == null || nums.length == 0) return;
        
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        for (int i = 0; i < nums.length; i++) {
            if (!map.containsKey(nums[i])) {
                map.put(nums[i], 1);
            } else {
                map.put(nums[i], map.get(nums[i]) + 1);
            }
        }
        
        int r = map.get(0) != null ? map.get(0) : 0;
        int w = map.get(1) != null ? map.get(1) : 0;
        int b = map.get(2) != null ? map.get(2) : 0;
        
        Color(nums, 0, 0, r);
        Color(nums, 1, r, w);
        Color(nums, 2, r + w, b);
    }

Follow up解法：

public void swap(int[] nums, int i, int j) {
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
    }
    
    public void sortColors(int[] nums) {
        if (nums == null || nums.length == 0) return;
        
        int r = 0, b = nums.length - 1;
        for (int i = 0; i < nums.length; i++) {
        //只要是遇到0或者2，就須要採起行動
            while (nums[i] == 0 && i >= r || (nums[i] == 2 && i <= b)) {
                if (nums[i] == 0) {
                    swap(nums, i, r++);
                } else {
                    swap(nums, i, b--);
                }
            }
        }
    }