一、今日知识点总结：

• 学会了二维数组的函数传参方式

二、今日做题记录：

1. 统计有序矩阵中的负数

int countNegatives(int** grid, int gridSize, int* gridColSize){
    int i, j, ans = 0;              
    int r = gridSize;
    int c = gridColSize[0];
    for(i = 0; i < r; ++i) {
        for(j = 0; j < c; ++j) {
            if(grid[i][j] < 0) {
                ++ans;
            } 
        }
    }
    return ans;
}

2. 矩阵对角线元素的和

int diagonalSum(int** mat, int matSize, int* matColSize){
    int r = matSize;
    int c = matColSize[0];
    int i;
    int ans = 0;
    for(i = 0; i < r; ++i) {
        ans += mat[i][i];
    }
    for(i = 0; i < r; ++i) {
        if(r-i-1 != i) {
            ans += mat[i][r-i-1];
        }
    }
    return ans;
}

3. 最富有客户的资产总量

int maximumWealth(int** accounts, int accountsSize, int* accountsColSize){
    int i, j;
    int maxv = -1, maxIdx, sumv;
    for(i = 0; i < accountsSize; ++i) {
        sumv = 0;
        for(j = 0; j < *accountsColSize; ++j) {
            sumv += accounts[i][j];
        }
        if(sumv > maxv) {
            maxv = sumv;
            maxIdx = i;
        }
    }
    return maxv;
}

4. 托普利茨矩阵

int checkSame(int** matrix, int sr, int sc, int maxr, int maxc) {
    int step = 0;
    while(1) {
        if(sr + step >= maxr) {
            break;
        }
        if(sc + step >= maxc) {
            break;
        }
        if(matrix[sr+step][sc+step] != matrix[sr][sc]) {
            return false;
        }
        ++step;
    }
    return true;
}

bool isToeplitzMatrix(int** matrix, int matrixSize, int* matrixColSize){
    int r = matrixSize;
    int c = matrixColSize[0];
    int i;
    for(i = 0; i < c; ++i) {
        if( !checkSame(matrix, 0, i, r, c) ) {
            return false;  
        }        
    }
    for(i = 0; i < r; ++i) {
        if( !checkSame(matrix, i, 0, r, c) ) {
            return false;
        }    
    }
    return true;
}

5. 矩阵中的幸运数

int min(int a, int b) {
    return a < b ? a : b;
}
int max(int a, int b) {
    return a > b ? a : b;
}

int* luckyNumbers (int** matrix, int matrixSize, int* matrixColSize, int* returnSize){
    int i, j;
    int r = matrixSize;
    int c = matrixColSize[0];
    int rmin[100];
    int cmax[100];
    int *ret = (int *)malloc( sizeof(int) * r * c );

    for(i = 0; i < r; ++i) {
        rmin[i] = 1000000;
        for(j = 0; j < c; ++j) {
            rmin[i] = min(rmin[i], matrix[i][j]);
        }
    }
    for(j = 0; j < c; ++j) {
        cmax[j] = 0;
        for(i = 0; i < r; ++i) {
            cmax[j] = max(cmax[j], matrix[i][j]);
        }
    }
    *returnSize = 0;
    for(i = 0; i < r; ++i) {
        for(j = 0; j < c; ++j) {
            if(matrix[i][j] == rmin[i] && matrix[i][j] == cmax[j]) {
                ret[ (*returnSize)++ ] = matrix[i][j];
            }
        }
    }
    return ret;
}