[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[0].empty()) return {};
        vector<vector<int>> sums = matrix;
        int m = matrix.size(), n = matrix[0].size();
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                int t = sums[i][j];
                if (i > 0) t += sums[i - 1][j];
                if (j > 0) t += sums[i][j - 1];
                if (i > 0 && j > 0) t -= sums[i - 1][j - 1];
                sums[i][j] = t;
                for (int p = 0; p <= i; ++p) {
                    for (int q = 0; q <= j; ++q) {
                        int d = sums[i][j];
                        if (p > 0) d -= sums[p - 1][j];
                        if (q > 0) d -= sums[i][q - 1];
                        if (p > 0 && q > 0) d += sums[p - 1][q - 1];
                        if (d == 0) 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[0].empty()) return {};
        int m = matrix.size(), n = matrix[0].size();
        for (int i = 0; i < n; ++i) {
            vector<int> sums(m, 0);
            for (int j = i; j < n; ++j) {
                for (int k = 0; k < m; ++k) {
                    sums[k] += matrix[k][j];
                }
                int curSum = 0;
                unordered_map<int, int> map{{0,-1}};
                for (int k = 0; k < m; ++k) {
                    curSum += sums[k];
                    if (map.count(curSum)) return {{map[curSum] + 1, i}, {k, j}};
                    map[curSum] = k;
                }
            }
        }
        return {};
    }
};

 

参考资料：

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