矩阵对角线元素的和 - 简单

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c++

topic: 1572. 矩阵对角线元素的和 - 力扣（LeetCode）

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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.

int w = 13;
int t = 38;

make a one-dimension vector.

// 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. 

// 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.

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.

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.

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;
    }
};