[LeetCode] Max Sum of Rectangle No Larger Than K 最大矩陣和不超過K
Given a non-empty 2D matrix matrix and an integer k, find the max sum of a rectangle in the matrix such that its sum is no larger than k.
Example:
Given matrix = [ [1, 0, 1], [0, -2, 3] ] k = 2
The answer is 2
. Because the sum of rectangle [[0, 1], [-2, 3]]
is 2 and 2 is the max number no larger than k (k = 2).
Note:
- The rectangle inside the matrix must have an area > 0.
- What if the number of rows is much larger than the number of columns?
Credits:
Special thanks to @fujiaozhu for adding this problem and creating all test cases.
這道題給了我們一個二維陣列,讓我們求和不超過的K的最大子矩形,那麼我們首先可以考慮使用brute force來解,就是遍歷所有的子矩形,然後計算其和跟K比較,找出不超過K的最大值即可。就算是暴力搜尋,我們也可以使用優化的演算法,比如建立累加和,參見之前那道題
解法一:
class Solution { public: int maxSumSubmatrix(vector<vector<int>>& matrix, int k) { if (matrix.empty() || matrix[0].empty()) return 0; int m = matrix.size(), n = matrix[0].size(), res = INT_MIN; int sum[m][n]; for (int i = 0; i < m; ++i) { for (int j = 0; j < n; ++j) { int t = matrix[i][j]; if (i > 0) t += sum[i - 1][j]; if (j > 0) t += sum[i][j - 1]; if (i > 0 && j > 0) t -= sum[i - 1][j - 1]; sum[i][j] = t; for (int r = 0; r <= i; ++r) { for (int c = 0; c <= j; ++c) { int d = sum[i][j]; if (r > 0) d -= sum[r - 1][j]; if (c > 0) d -= sum[i][c - 1]; if (r > 0 && c > 0) d += sum[r - 1][c - 1]; if (d <= k) res = max(res, d); } } } } return res; } };
下面這個演算法進一步的優化了執行時間,這個演算法是基於計算二維陣列中最大子矩陣和的演算法,可以參見youtube上的這個視訊Maximum Sum Rectangular Submatrix in Matrix dynamic programming/2D kadane。這個演算法巧妙在把二維陣列按行或列拆成多個一維陣列,然後利用一維陣列的累加和來找符合要求的數字,這裡用了lower_bound來加快我們的搜尋速度,也可以使用二分搜尋法來替代。我們建立一個集合set,然後開始先放個0進去,為啥要放0呢,因為我們要找lower_bound(curSum - k),當curSum和k相等時,0就可以被返回了,這樣我們就能更新結果了。由於我們對於一維陣列建立了累積和,那麼sum[i,j] = sum[i] - sum[j],其中sums[i,j]就是目標子陣列需要其和小於等於k,然後sums[j]是curSum,而sum[i]就是我們要找值,當我們使用二分搜尋法找sum[i]時,sum[i]的和需要>=sum[j] - k,所以也可以使用lower_bound來找,參見程式碼如下:
解法二:
class Solution { public: int maxSumSubmatrix(vector<vector<int>>& matrix, int k) { if (matrix.empty() || matrix[0].empty()) return 0; int m = matrix.size(), n = matrix[0].size(), res = INT_MIN; for (int i = 0; i < n; ++i) { vector<int> sum(m, 0); for (int j = i; j < n; ++j) { for (int k = 0; k < m; ++k) { sum[k] += matrix[k][j]; } int curSum = 0, curMax = INT_MIN; set<int> s; s.insert(0); for (auto a : sum) { curSum += a; auto it = s.lower_bound(curSum - k); if (it != s.end()) curMax = max(curMax, curSum - *it); s.insert(curSum); } res = max(res, curMax); } } return res; } };
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