矩阵对角线元素的和 - 简单

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

c++

topic: 1572. 矩阵对角线元素的和 - 力扣(LeetCode)

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

Look at the problems immediately.

|----------------------------------------------------------------------------|
| |

vector<vector<int>>& mat means mat is a two-dimension vector. Let's review the basic usage of the creating vector in c++.

make an integer.

cpp 复制代码
int w = 13;
int t = 38;

make a one-dimension vector.

cpp 复制代码
// 直接给定数组,数组的名字是自定义的
vector<int> w = {1, 3, 3, 8};

// 构造一个数组,包含13个元素,每个元素是 38
vector<int> t(13, 38);

make a two-dimension vector. And when talks about two-dimension vector, it is made of many one-dimension vctors.

cpp 复制代码
// 一维数组
vecotr<int> w(13, 38);

输出:
38 38 38 38 38 38 38 38 38 38 38 38 38


// 二维数组就是规定了有几个一维数组、
vector<vector<int>> t(13, vector<int>(13, 38));

输出:
38 38 38 38 38 38 38 38 38 38 38 38 38
38 38 38 38 38 38 38 38 38 38 38 38 38
38 38 38 38 38 38 38 38 38 38 38 38 38
38 38 38 38 38 38 38 38 38 38 38 38 38
38 38 38 38 38 38 38 38 38 38 38 38 38
38 38 38 38 38 38 38 38 38 38 38 38 38
38 38 38 38 38 38 38 38 38 38 38 38 38
38 38 38 38 38 38 38 38 38 38 38 38 38
38 38 38 38 38 38 38 38 38 38 38 38 38
38 38 38 38 38 38 38 38 38 38 38 38 38
38 38 38 38 38 38 38 38 38 38 38 38 38
38 38 38 38 38 38 38 38 38 38 38 38 38
38 38 38 38 38 38 38 38 38 38 38 38 38

I like the basic usages of everything so much. Making full usage of the things keeps claen. Many people want to learn too much skills, which I think donnot have to. Keep things simple.

I think when looking at the mat, getting the size is always first.

cpp 复制代码
class Solution {
public:
    int diagonalSum(vector<vector<int>>& mat) {
        
        int n = mat.size(); // get the size of mat
        int sum = 0;
    }
};

mat[a][b] means the element lies in line a column b.

cpp 复制代码
class Solution {
public:
    int diagonalSum(vector<vector<int>>& mat) {
        
        int n = mat.size(); // get the size of mat
        int sum = 0;

        for (int i = 0; i < n; i++)
        {
            sum = sum + mat[i][i];
            sum = sum + mat[i][n - 1 - i];
        }

        return sum;
    }
};

|----------------------------------------------------------------------------|
| |

This problen is easy but sumething wrong. Soon I find the key point. 5 is really a special one. It lies in both main diagonal and counter diagonal.

|----------------------------------------------------------------------------|
| |

just minus it.

cpp 复制代码
class Solution {
public:
    int diagonalSum(vector<vector<int>>& mat) {
        
        int n = mat.size(); // get the size of mat
        int sum = 0;

        for (int i = 0; i < n; i++)
        {
            sum = sum + mat[i][i];
            sum = sum + mat[i][n - 1 - i];
        }

        // 如果奇数个元素,那么得减掉正中心的元素,因为他被计算了两遍
        if (n % 2 == 1)
        {
            sum = sum - mat[(n - 1) / 2][(n - 1) / 2];
        }

        return sum;
    }
};

|----------------------------------------------------------------------------|
| |

相关推荐
打螺丝否7 小时前
稠密矩阵和稀疏矩阵的对比
python·机器学习·矩阵
人机与认知实验室7 小时前
人机环境系统智能矩阵理论
线性代数·矩阵
fFee-ops2 天前
73. 矩阵置零
线性代数·矩阵
星逝*2 天前
LeetCode刷题-top100( 矩阵置零)
算法·leetcode·矩阵
码界奇点2 天前
豆包新模型矩阵与PromptPilot构建企业级AI开发的体系化解决方案
人工智能·线性代数·ai·语言模型·矩阵·硬件工程
酸奶乳酪2 天前
矩阵和向量的双重视角
线性代数·矩阵
阿维的博客日记2 天前
LeetCode 240: 搜索二维矩阵 II - 算法详解(秒懂系列
算法·leetcode·矩阵
桐果云3 天前
解锁桐果云零代码数据平台能力矩阵——赋能零售行业数字化转型新动能
大数据·人工智能·矩阵·数据挖掘·数据分析·零售
自信的小螺丝钉3 天前
Leetcode 240. 搜索二维矩阵 II 矩阵 / 二分
算法·leetcode·矩阵
lytk993 天前
矩阵中寻找好子矩阵
线性代数·算法·矩阵