成品网站源码1688的优势百度的推广方式有哪些
矩阵其实就是二维数组，这些题目在9日集训中已经做过，这里做的方法大致相同。
第一题 1351. 统计有序矩阵中的负数
int countNegatives(int** grid, int gridSize, int* gridColSize) {int r = gridSize;int c = gridColSize[0];int ret = 0;for(int i = 0; i < r; ++i) {for(int j = 0; j < c; ++j) {if(grid[i][j] < 0) ret++;}}return ret;
}
第二题 1672. 最富有客户的资产总量
int maximumWealth(int** accounts, int accountsSize, int* accountsColSize) {int r = accountsSize;int c = accountsColSize[0];int max = 0;for(int i = 0; i < r; ++i) {int temp = 0;for(int j = 0; j < c; ++j) {temp += accounts[i][j];}if(temp > max) max = temp;}return max;
}
第三题 1572. 矩阵对角线元素的和
int diagonalSum(int** mat, int matSize, int* matColSize){//取矩阵行和列长int r = matSize;int sum = 0;for(int i = 0; i < r; ++i) {sum += mat[i][i];}int index = 0;for(int i = r - 1; i >= 0; --i) {sum += mat[index++][i];}if(r % 2 == 1) sum -= mat[r / 2][r / 2];return sum;
}
第四题 2643. 一最多的行
/*** Note: The returned array must be malloced, assume caller calls free().*/
int* rowAndMaximumOnes(int** mat, int matSize, int* matColSize, int* returnSize){int * ret = (int *)malloc(sizeof(int) * 2);ret[0] = 0;ret[1] = 0;int r = matSize;int c = matColSize[0];for(int i = 0; i < r; ++i) {int sum1 = 0;for(int j = 0; j < c; ++j) {if(mat[i][j] == 1) {sum1++;}}if(sum1 > ret[1]) {ret[0] = i;ret[1] = sum1;}}*returnSize = 2;return ret;
}