轮转数组 - 中等-优快云博客

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

C++

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

Give the topic a inspection.

This topic looks like an easy one and just make another vector to store the changed vector.

class Solution {
public:
    void rotate(vector<int>& nums, int k) {
        
        int n = nums.size();
    }
};

And then iterate through the entire vector. Before that, I really enjoy using the basic usages in vector. Here are some basic usages of vector in c++.

create and initialise

vector<int> v1;                // 空 → []
vector<int> v2(3);             // 3个0 → [0,0,0]
vector<int> v3 = {1, 2, 3};    // → [1,2,3]
vector<string> v4(2, "A");     // → ["A","A"]

add some elements

vector<int> v = {1, 2};
v.push_back(3);        // → [1,2,3]   （尾部插入）
v.emplace_back(4);     // → [1,2,3,4] （更高效）
v.insert(v.begin(), 0);// → [0,1,2,3,4] （开头插入）
v.insert(v.begin()+2, 9); // → [0,1,9,2,3,4] （第3位插入）


vector<int> v = {1, 2, 3, 4, 5, 6, 7};
// 在开头插入最后3个元素（5,6,7）
v.insert(v.begin(), v.end() - 3, v.end()); // 输出：5 6 7 1 2 3 4 5 6 7

delete some elements

vector<int> v = {10,20,30,40};
v.pop_back();         // → [10,20,30]（删除尾部）
v.erase(v.begin()+1); // → [10,30]   （删除第2个元素）
v.clear();            // → []        （清空）

visit the element

vector<int> v = {5,6,7};
int a = v[0];         // a=5（不检查越界）
int b = v.at(1);      // b=6（越界会抛异常）
int c = v.front();    // c=5（首元素）
int d = v.back();     // d=7（尾元素）

change the size of the vector

vector<int> v = {1,2};
v.resize(4);          // → [1,2,0,0]（填充0到4个元素）
v.resize(3);          // → [1,2,0] （截断到3个）
v.reserve(10);        // 容量变为10，但元素不变

This is almost the 5 time I learn how to do the basic usages of the vector.

class Solution {
public:
    void rotate(vector<int>& nums, int k) {

        nums.insert(nums.begin(), nums.end() - k, nums.end());
        nums.erase(nums.end() - k, nums.end());
      
    }
};

add something.

class Solution {
public:
    void rotate(vector<int>& nums, int k) {

        int n = nums.size();
        if (k > n) 
        {
            k = k % n;
        }

        nums.insert(nums.begin(), nums.end() - k, nums.end());
        nums.erase(nums.end() - k, nums.end());
    }
};

traditional manipulate: debug.

步骤	当前执行代码行	变量状态变化	条件判断	循环迭代	输出内容	操作注释
1	`void rotate(...)`	`nums` = [1,2,3], `k` = 1	-	-	-	进入函数
2	`int n = nums.size()`	`n` = 3 (原数组长度)	-	-	-	获取数组长度
3	`if (k > n)`	`k`(1) ≤ `n`(3) → false	不执行if块	-	-	检查k是否需要取模
4	`nums.insert(...)`	`nums` = [1,2,3] → [3,1,2,3]	-	-	-	插入后k个元素到开头
5	`nums.erase(...)`	`nums` = [3,1,2,3] → [3,1,2]	-	-	-	删除最后k个元素
6	函数结束	`nums` = [3,1,2]	-	-	-	返回修改后的数组

my code is right.