移动零 - 简单

*************

C++

topic: 283. 移动零 - 力扣（LeetCode）

*************

In the past few days, I did the shadowing reading practice to be a native speaker. It is really annoyiny that the hoster speeks too fast and I cannot keep up with the speed. One thing that really upset is that I donot do the practices a few days later. It seems working and then I give up. I realize that it is not good habbit and need to change it. After a half-year self-learning c++, it is the time to relearn the code to find something new. 

编程基础 0 到 1 - 学习计划 - 力扣（LeetCode）全球极客挚爱的技术成长平台https://leetcode.cn/studyplan/programming-skills/

OK, have a glance at todays topic:

This time I want to write the code with personal style. Good habbit maks every steps.

If nums[1] = 0, nums[2] = 1, swap the nums. nums[1] = 1, nums[2] = 0. This is the easy-understand way to line the 0 elements in the end of the strings. If muns[right] != 0, swap the left element and the right element.

class Solution {
public:
    void moveZeroes(vector<int>& nums) {

        private:

        public :

        if (nums[right] != 0)
        {
            swap(nums[left], nums[right]);
        }

        
    }
};

the grammar of the swap is easy.

// swap 是通过三次赋值实现：
void swap(T& a, T& b) 
{
    T temp = a;
    a = b;
    b = temp;
}

then move the pointer right.

class Solution {
public:
    void moveZeroes(vector<int>& nums) {

        private:

        public :

        if (nums[right] != 0)
        {
            // 交换左右元素，确保右边的元素为0
            swap(nums[left], nums[right]);
            ++left;
        }

        
    }
};

make sure that right and left strat from 0, so the checking will travel all the elements.

class Solution {
public:
    void moveZeroes(vector<int>& nums) {

        private:

        public :

        if (nums[right] != 0)
        {
            for (right = 0; right < n; ++right)
            {
            // 交换左右元素，确保右边的元素为0
            swap(nums[left], nums[right]);
            ++left;
            }
            
        }

        
    }
};

and then declare the variables.

class Solution {
public:
    void moveZeroes(vector<int>& nums) {

        private:
        int left = 0;
        int right = 0;
        int n = nums.size();

        public :
        if (nums[right] != 0)
        {
            for (right = 0; right < n; ++right)
            {
            // 交换左右元素，确保右边的元素为0
            swap(nums[left], nums[right]);
            ++left;
            }
            
        }

        
    }
};

In the fact, the code is wrong even if the steps is right. Because it is already have a class.

so in the class Solution, donot need a private to decalare the variables any more.

class Solution {
public:
    void moveZeroes(vector<int>& nums) {

        //private:
        int left = 0;
        //int right = 0;
        int n = nums.size();

        //public :
        
        {
            for (right = 0; right < n; ++right)
            {
                if (nums[right] != 0)
                {
                   // 交换左右元素，确保右边的元素为0
                   swap(nums[left], nums[right]);
                   ++left;
                } 
            }  
        } 
    }
};

 

Even if the code above is wrong. The element right is not declared.

class Solution {
public:
    void moveZeroes(vector<int>& nums) {

        //private:
        int left = 0;
        //int right = 0;
        int n = nums.size();

        //public :
        
        {
            // 这里要定义 right 
            for (int right = 0; right < n; ++right)
            {
                if (nums[right] != 0)
                {
                   // 交换左右元素，确保右边的元素为0
                   swap(nums[left], nums[right]);
                   ++left;
                } 
            }  
        } 
    }
};

Here annotate the int right = 0 in the line 7 is necessary. otherwise the int right is always 0.

the code start from line 1 to the line end. int right changes should happen in the if loop, which makes code runs well.

class Solution {
public:
    void moveZeroes(vector<int>& nums) {

        // declare the the varables
        int left = 0;
        int n = nums.size();

        // the main code
        for (int right = 0; right < n; ++right) 
        {
            if (nums[right] != 0) 
            {
                // 交换左右元素，确保右边的元素为0
                swap(nums[left], nums[right]);
                ++left;
            }
        }
    }
};

debug the code is a really funny thing in vscode, but here cannot present the process of the debug. So list the process of the running.

---

‌初始状态‌

• ‌数组‌：[0, 1, 0, 3, 12]
• ‌指针位置‌：left = 0，right = 0‌

---

‌分步执行‌

1. ‌Step 1（right=0）‌

   • ‌检测元素‌：nums[right] = 0
   • ‌操作‌：跳过交换，right 右移至 1‌
   • ‌当前数组‌：[0, 1, 0, 3, 12]
     （left=0，right=1）

2. ‌Step 2（right=1）‌

   • ‌检测元素‌：nums[right] = 1（非零）
   • ‌交换操作‌：
     • 交换 nums[left]（0）与 nums[right]（1）‌
     • left 右移至 1‌
   • ‌交换后数组‌：[1, 0, 0, 3, 12]
     （left=1，right=2）

3. ‌Step 3（right=2）‌

   • ‌检测元素‌：nums[right] = 0
   • ‌操作‌：跳过交换，right 右移至 3‌
   • ‌当前数组‌：[1, 0, 0, 3, 12]
     （left=1，right=3）

4. ‌Step 4（right=3）‌

   • ‌检测元素‌：nums[right] = 3（非零）
   • ‌交换操作‌：
     • 交换 nums[left]（0）与 nums[right]（3）‌
     • left 右移至 2‌
   • ‌交换后数组‌：[1, 3, 0, 0, 12]
     （left=2，right=4）

5. ‌Step 5（right=4）‌

   • ‌检测元素‌：nums[right] = 12（非零）
   • ‌交换操作‌：
     • 交换 nums[left]（0）与 nums[right]（12）‌
     • left 右移至 3‌
   • ‌最终数组‌：[1, 3, 12, 0, 0]
     （left=3，right=5）