75. Sort Colors

longmon 發(fā)布于2019-08-14 14:09 / 1190人閱讀

摘要：題目解題可以參考的這題比較簡單，就沒有用書里的解法，的思想就是交換，既然只能，那就一次至少搞定一個數(shù)啦解法解法只要是遇到或者，就需要采取行動

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：
可以參考EPI的14.8, 這題比較簡單，就沒有用書里的解法，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(shù)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 map = new HashMap();
        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，就需要采取行動
            while (nums[i] == 0 && i >= r || (nums[i] == 2 && i <= b)) {
                if (nums[i] == 0) {
                    swap(nums, i, r++);
                } else {
                    swap(nums, i, b--);
                }
            }
        }
    }