【Medium】75. Sort Colors

最新推荐文章于 2020-03-23 14:48:14 发布

原创最新推荐文章于 2020-03-23 14:48:14 发布 · 258 阅读

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本文介绍了一种不使用库排序函数的三色排序算法，通过两种方法实现数组中红、白、蓝三种颜色对象的排序。第一种方法采用计数排序，统计每种颜色的数量后按顺序填充数组；第二种方法采用单次遍历，使用两个指针进行原地排序。

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.

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?

解法一

中心思想：

很简单地遍历一边数组，记录每个数字出现的次数，然后依次打印出来

class Solution {
public:
    void sortColors(vector<int>& nums) {
        int n = nums.size();
        if (n == 0)
            return;
        
        int zero = 0, one = 0, two = 0;
        
        for (int i = 0; i < n; i++){
            if (nums[i] == 0){
                zero++;
            }
            else if (nums[i] == 1){
                one++;
            }
            else if (nums[i] == 2){
                two++;
            }
        }
        
        for (int i = 0; i < zero; i++){
            nums[i] = 0;
        }
        for (int i = zero; i < zero + one; i++){
            nums[i] = 1;
        }
        for (int i = zero + one; i < zero+one+two; i++){
            nums[i] = 2;
        }
        
        return;
        
    }
};

解法二

中心思想：

用两个指针left, right来分割数组。left用来记录第一个1， right用来记录从右数起第一个非2数。Left起始为0，right起始为n-1. 遍历时，遇到0，就将nums[left]和nums[i]换一换，将0换到左边去，由于现在nums[left]是0了，left++, i++;遇到1，left和right都不动，i++; 遇到2，和Nums[right]换一换把2换到右边去，现在Nums[right]上是2所以right--, nums[i]上是0或者1，所以i不动，要继续检查是0还是1.

class Solution {
public:
    void sortColors(vector<int>& nums) {
<span style="white-space:pre">	</span>int n = nums.size();
        if (n == 0)
            return;
        
        int left = 0, right = n-1, i = 0;
        
        while (i <= right){
            if (nums[i] ==0){
                swap(nums[left], nums[i]);
                i++;
                left++;
            }
            else if (nums[i] == 1){
                i++;
            }
            else{
                swap(nums[right], nums[i]);
                right--;
            }
        }
        
        return;
    }
};