前言

hello 大家好啊，今天是第七天啦，加油加油。

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1351. 统计有序矩阵中的负数

class Solution {public:    int countNegatives(vector<vector<int>>& grid) { int rows = grid.size(), cols = grid[0].size(); int ans = 0; for(int i = rows-1; i >= 0; --i) {     for(int j = cols-1; j >= 0; --j)     {  if(grid[i][j] < 0)      ++ans;     } } return ans;    }};

1572. 矩阵对角线元素的和

class Solution {public:    int diagonalSum(vector<vector<int>>& mat) { int n = mat.size(); int sum = 0; for(int i = 0; i < n; ++i) {     sum += mat[i][i]; } for(int i = 0; i < n; ++i) {     if(n-i-1 != i)  sum += mat[i][n-i-1]; } return sum;    }};

1672. 最富有客户的资产总量

单独统计每一行的再比较就行

class Solution {public:    int maximumWealth(vector<vector<int>>& accounts) { int maxSum = 0; for(int i = 0;i < accounts.size(); ++i) {     int sum = 0;     for(int j = 0; j < accounts[i].size(); ++j)     {  sum += accounts[i][j];     }     maxSum = max(maxSum, sum); } return maxSum;    }};

766. 托普利茨矩阵

class Solution {public:    bool isToeplitzMatrix(vector<vector<int>>& matrix) { int rows = matrix.size(), cols = matrix[0].size(); if(rows == 1 || cols == 1)     return true; for(int i = 0; i < rows-1; ++i) {     for(int j = 0; j < cols-1; ++j)     {  if(matrix[i][j] != matrix[i+1][j+1])      return false;     } } return true;    }};

1380. 矩阵中的幸运数

class Solution {public:    vector<int> luckyNumbers (vector<vector<int>>& matrix) { int rows = matrix.size(), cols = matrix[0].size(); //开2个数组，分别存每一行的最小值，每一列的最大值 vector<int> row(rows, 0); vector<int> col(cols, 0); vector<int> ret; for(int i = 0; i < rows; ++i) {     for(int j = 0; j < cols; ++j)     {  if(matrix[i][j] < matrix[i][row[i]])      row[i] = j;  if(matrix[i][j] > matrix[col[j]][j])      col[j] = i;     } } for(int i = 0; i < rows; ++i) {     if(i == col[row[i]])     {  ret.push_back(matrix[i][row[i]]);     } } return ret;    }};

1582. 二进制矩阵中的特殊位置

/*- 计数：即每一行1的个数 和 每一列1的一个- 再次遍历一旦出现1，且对应的行和列数量都是1，就是结果，累加*/class Solution {public:    int numSpecial(vector<vector<int>>& mat) { int rows = mat.size(); int cols = mat[0].size(); int rowCnt[rows]; int colCnt[cols]; memset(rowCnt, 0, sizeof(rowCnt)); memset(colCnt, 0, sizeof(colCnt)); for (int i = 0; i < rows; ++i) {     for (int j = 0; j < cols; ++j)     {  int curr = mat[i][j] == 1;  rowCnt[i] += curr;  colCnt[j] += curr;     } } int res = 0; for (int i = 0; i < rows; ++i) {     for (int j = 0; j < cols; ++j)     {  if (mat[i][j] == 1 && rowCnt[i] == 1 && colCnt[j] == 1)  {      ++res;  }     } } return res;    }};

463. 岛屿的周长

class Solution {    constexpr static int dx[4] = {0, 1, 0, -1};    constexpr static int dy[4] = {1, 0, -1, 0};public:    int islandPerimeter(vector<vector<int>> &grid) { int n = grid.size(), m = grid[0].size(); int ans = 0; for (int i = 0; i < n; ++i) {     for (int j = 0; j < m; ++j) {  if (grid[i][j]) {      int cnt = 0;      for (int k = 0; k < 4; ++k) {   int tx = i + dx[k];   int ty = j + dy[k];   if (tx < 0 || tx >= n || ty < 0 || ty >= m || !grid[tx][ty]) {cnt += 1;   }      }      ans += cnt;  }     } } return ans;    }};

尾声

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👀👀由于笔者水平有限，在今后的博文中难免会出现错误之处，本人非常希望您如果发现错误，恳请留言批评斧正，希望和大家一起学习，一起进步ヽ(￣ω￣(￣ω￣〃)ゝ，期待您的留言评论。
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