【一天一道LeetCode】#75. Sort Colors

一天一道LeetCode

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（一）题目

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.

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?

（二）解题

题目大意：数组包含0,1,2，按照大小排列这些数字。

1、最简单的方法

利用STL的sort一句话就解决了。

class Solution {
public:
    void sortColors(vector<int>& nums) {
        sort(nums.begin(),nums.end());
    }
};

2、两个指针

算法思想很简单，把所有的0交换到队首，把所有的1交换到队尾

class Solution {
public:
    void sortColors(vector<int>& nums) {
        int start = 0;
        int end = nums.size()-1;
        for(int i = 0 ; i<nums.size();i++)//把0交换到前面
        {
            if(nums[i]==0) {
                if(i!=start) swap(nums[i],nums[start]);
                start++;
            }
        }
        for(int i = nums.size()-1 ; i>=start ;i--)//把2交换到尾部
        {
            if(nums[i]==2){
                if(i!=end) swap(nums[i],nums[end]);
                end--;
            }
        }
    }
};

3、两个指针优化版

维持三个指针i、j和k，从0~i表示1，i+1~j表示1，k~nums.size()表示2

代码也比较简单：

class Solution {
public:
    void sortColors(vector<int>& nums) {
        int i = 0, j = i, k = nums.size() - 1;
        while(j <= k){
            if(nums[j] == 0) swap(nums[i++], nums[j++]);//0交换到前面
            else if(nums[j] == 1) j++;//1保持不动
            else swap(nums[k--], nums[j]);//2交换到尾部
        }
    }
};