C 解法, 执行用时: 55ms, 内存消耗: 7556KB, 提交时间: 2022-07-18

int brickwall(int** wall, int wallRowLen, int* wallColLen ) {
    int h = 0;
    for (int j = 0; j < wallColLen[0]; j++) {
        h += wall[0][j];
    }
    
    int a[h + 1];
    memset(a, 0, sizeof(a));   
    for (int i = 0; i < wallRowLen; i++) {
        h = 0;
        for (int j = 0; j < wallColLen[i]; j++) {
            for (int k = 1; k < wall[i][j]; k++) {
                a[h+k] += 1;
            }
            h += wall[i][j];
        }
    }
    
    int z = wallRowLen;
    for (int i = 1; i < h; i++) {
        if (a[i] < z) {
            z = a[i];
        }
    }    
    return z;
}

C 解法, 执行用时: 57ms, 内存消耗: 7436KB, 提交时间: 2022-06-28

/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param wall int整型二维数组 
 * @param wallRowLen int wall数组行数
 * @param wallColLen int* wall数组列数
 * @return int整型
 *
 * C语言声明定义全局变量请加上static，防止重复定义
 */
int brickwall(int** wall, int wallRowLen, int* wallColLen ) {
    int width = 0;
    for (int j = 0; j < wallColLen[0]; j++) {
        width += wall[0][j];
    }
    
    int hash[width + 1];
    memset(hash, 0, sizeof(hash));
    
    for (int i = 0; i < wallRowLen; i++) {
        width = 0;
        for (int j = 0; j < wallColLen[i]; j++) {
            for (int k = 1; k < wall[i][j]; k++) {
                hash[width+k] += 1;
            }
            width += wall[i][j];
        }
    }
    
    int minCount = wallRowLen;
    for (int i = 1; i < width; i++) {
        if (hash[i] < minCount) {
            minCount = hash[i];
        }
    }
    
    return minCount;
}

C++ 解法, 执行用时: 57ms, 内存消耗: 7664KB, 提交时间: 2022-07-18

class Solution {
  public:
    int brickwall(vector<vector<int> >& wall) {
        int a = 0;
        for (auto f : wall[0]) {
            a += f;
        }

        vector<int> v(a + 1, 0);
        for (int i = 0; i < wall.size(); i ++) {
            int b = 0;
            for (int j = 0; j < wall[i].size(); j ++) {
                b += wall[i][j];
                v[b]++;
            }
        }

        int h = wall.size();
        int z = INT_MAX;
        for (int i = 1; i <= a - 1; i ++) {
            z = min(z, h - v[i]);
        }
        return z;
    }
};

C++ 解法, 执行用时: 58ms, 内存消耗: 7668KB, 提交时间: 2022-08-06

class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param wall int整型vector<vector<>> 
     * @return int整型
     */
    int brickwall(vector<vector<int> >& wall) {
        // write code here
        int a=0;
        for(auto f:wall[0])
        {
            a+=f;
        }
        vector<int> v(a+1,0);
        for(int i=0;i<wall.size();i++)
        {
            int b=0;
            for(int j=0;j<wall[i].size();j++)
            {
                b+=wall[i][j];
                v[b]++;
            }
        }
        int h=wall.size();
        int z=INT_MAX;
        for(int i=1;i<=a-1;i++)
        {
            z=min(z,h-v[i]);
        }
        return z;
    }
};

C++ 解法, 执行用时: 58ms, 内存消耗: 7668KB, 提交时间: 2022-07-03

class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param wall int整型vector<vector<>> 
     * @return int整型
     */
    int brickwall(vector<vector<int> >& wall) {
        // write code here
        int n = 0;
        for(auto subvecn : wall[0]){
            n += subvecn;
        }
        vector<int> wallline(n + 1, 0);
        for(int i = 0; i < wall.size(); i ++){
            int len = 0;
            for(int j = 0; j < wall[i].size(); j ++){
                len += wall[i][j];
                wallline[len]++;
            }
        }
        int layer = wall.size();
        int res = INT_MAX;
        for(int i = 1; i <= n - 1; i ++){
            res = min(res, layer - wallline[i]);
        }
        return res;
    }
};