[LintCode] Submatrix Sum 子矩阵之和

Given an integer matrix, find a submatrix where the sum of numbers is zero. Your code should return the coordinate of the left-up and right-down number.

Have you met this question in a real interview?

Yes

Example

Given matrix

[
  [1 ,5 ,7],
  [3 ,7 ,-8],
  [4 ,-8 ,9],
]

return [(1,1), (2,2)]

Challenge

O(n3) time.

这道题跟LeetCode上的那道Max Sum of Rectangle No Larger Than K很类似。

解法一：

class Solution {
public:
    /**
     * @param matrix an integer matrix
     * @return the coordinate of the left-up and right-down number
     */
    vector<vector<int>> submatrixSum(vector<vector<int>>& matrix) {
        if (matrix.empty() || matrix[].empty()) return {};
        vector<vector<int>> sums = matrix;
        int m = matrix.size(), n = matrix[].size();
        for (int i = ; i < m; ++i) {
            for (int j = ; j < n; ++j) {
                int t = sums[i][j];
                if (i > ) t += sums[i - ][j];
                if (j > ) t += sums[i][j - ];
                if (i >  && j > ) t -= sums[i - ][j - ];
                sums[i][j] = t;
                for (int p = ; p <= i; ++p) {
                    for (int q = ; q <= j; ++q) {
                        int d = sums[i][j];
                        if (p > ) d -= sums[p - ][j];
                        if (q > ) d -= sums[i][q - ];
                        if (p >  && q > ) d += sums[p - ][q - ];
                        if (d == ) return {{p, q}, {i, j}};
                    }
                }
            }
        }
        printVec(sums);
        return {};
    }
};

解法二：

class Solution {
public:
    /**
     * @param matrix an integer matrix
     * @return the coordinate of the left-up and right-down number
     */
    vector<vector<int>> submatrixSum(vector<vector<int>>& matrix) {
        if (matrix.empty() || matrix[].empty()) return {};
        int m = matrix.size(), n = matrix[].size();
        for (int i = ; i < n; ++i) {
            vector<int> sums(m, );
            for (int j = i; j < n; ++j) {
                for (int k = ; k < m; ++k) {
                    sums[k] += matrix[k][j];
                }
                int curSum = ;
                unordered_map<int, int> map{{,-}};
                for (int k = ; k < m; ++k) {
                    curSum += sums[k];
                    if (map.count(curSum)) return {{map[curSum] + , i}, {k, j}};
                    map[curSum] = k;
                }
            }
        }
        return {};
    }
};

参考资料：

http://www.jiuzhang.com/solutions/submatrix-sum/