LeetCode: Sort Colors 三种颜色排序多排序算法

不念不忘少年蓝@ 2021-06-24 16:10 322阅读 0赞

**试题：**  
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?  
**代码：**  
o(n^2):

//冒泡排序：
    class Solution {
        public void sortColors(int[] nums) {
            for(int i=0; i<nums.length-1; i++){
                for(int j=0; j<nums.length-1-i; j++){
                    if(nums[j]>nums[j+1]){
                        int tmp = nums[j+1];
                        nums[j+1] = nums[j];
                        nums[j] = tmp;
                    }
                }
            }
        }
    }

//选择排序
    class Solution {
        public void sortColors(int[] nums) {
            for(int i=0; i<nums.length-1; i++){
                int minindex = i;
                for(int j=i+1; j<nums.length; j++){
                    if(nums[j]<nums[minindex]){
                        minindex = j;
                    }
                }
                int tmp = nums[minindex];
                nums[minindex] = nums[i];
                nums[i] = tmp;
            }
        }
    }

//插入排序
    class Solution {
        public void sortColors(int[] nums) {
            for(int i=1; i<nums.length; i++){
                int index = i-1;
                int tmp = nums[i];
                while(index>=0 && nums[index]>tmp){
                    nums[index+1] = nums[index];
                    index--;
                };
                nums[index+1] = tmp;
            }
        }
    }

o(nlogn):

//快速排序

o(n)：  
由于题中只有三种数值，那么对于遇到0时我们可以往低位置放置，遇到2时往高位置放置即可。

class Solution {
        public void sortColors(int[] nums) {
            int low=0, pos=0, high=nums.length-1;
            while(pos<=high){
                if(nums[pos]==0){
                    swap(nums, low, pos);
                    low++;pos++;
                }else if(nums[pos]==2){
                    swap(nums, high, pos);
                    high--;
                }else{
                    pos++;
                }
            }
        }
        
        private void swap(int[] nums, int i, int j){
            int tmp = nums[i];
            nums[i] = nums[j];
            nums[j] = tmp;
        }
    }