75. Sort Colors (Array)

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?

思路: 使用两个指针，分别指向最后一个红色的后一个位置，以及第一个蓝色蓝色的前一个位置。如果当前扫描到的元素是红色，那么与第一个指针互换，若为蓝色，与蓝色头部指针互换，若为白色，不做任何操作。

class Solution {
public:
    void sortColors(vector<int>& nums) {
        int n = nums.size();
        int redTail = 0;
        int blueHead = n-1;
        int tmp;
        
        for(int i = 0; i <= blueHead;){
            if(nums[i]==0){//当前是红色
                if(nums[redTail]==0){//红色尾指针指向红色（说明没有白色）
                    redTail++;
                    i++; //不处理当前元素
                    continue;
                }
                
                tmp = nums[redTail];
                nums[redTail] = nums[i];
                redTail++;
                if(tmp==1){ //红色尾指针指向白色
                    nums[i] = tmp; //交换红色和白色
                    i++;
                }
                else{ //红色尾指针指向蓝色（说明目前没有白色）
                    nums[i] = nums[blueHead]; //把blueHead处的元素放到i
                    nums[blueHead]=tmp; //把蓝色放到blueHead处
                    blueHead--;
                }
            }
            else if(nums[i]==1){//如果当前是白色，不处理当前元素
                i++;
            }
            else{ //如果当前是蓝色
                tmp = nums[blueHead];
                nums[blueHead] = nums[i];//把蓝色放到blueHead处
                nums[i] = tmp;//把blueHead处的元素放到i
                blueHead--;
            }
        }
    }
};

 

转载于:https://www.cnblogs.com/qionglouyuyu/p/4853021.html