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:
Input: matrix = [[1,0,1],[0,-2,3]], k = 2
Output: 2
Explanation: 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.
这道题给了我们一个二维数组,让求和不超过的K的最大子矩形,那么首先可以考虑使用 brute force 来解,就是遍历所有的子矩形,然后计算其和跟K比较,找出不超过K的最大值即可。就算是暴力搜索,也可以使用优化的算法,比如建立累加和,参见之前那道题 Range Sum Query 2D - Immutable,可以快速求出任何一个区间和,下面的方法就是这样的,当遍历到 (i, j) 时,计算 sum(i, j),表示矩形 (0, 0) 到 (i, j) 的和,然后遍历这个矩形中所有的子矩形,计算其和跟K相比,这样既可遍历到原矩形的所有子矩形,参见代码如下:
解法一:
class Solution {
public:
int maxSumSubmatrix(vector<vector<int>>& matrix, int k) {
if (matrix.empty() || matrix[].empty()) return ;
int m = matrix.size(), n = matrix[].size(), res = INT_MIN;
int sum[m][n];
for (int i = ; i < m; ++i) {
for (int j = ; j < n; ++j) {
int t = matrix[i][j];
if (i > ) t += sum[i - ][j];
if (j > ) t += sum[i][j - ];
if (i > && j > ) t -= sum[i - ][j - ];
sum[i][j] = t;
for (int r = ; r <= i; ++r) {
for (int c = ; c <= j; ++c) {
int d = sum[i][j];
if (r > ) d -= sum[r - ][j];
if (c > ) d -= sum[i][c - ];
if (r > && c > ) d += sum[r - ][c - ];
if (d <= k) res = max(res, d);
}
}
}
}
return res;
}
};
下面这个算法进一步的优化了运行时间,这个算法是基于计算二维数组中最大子矩阵和的算法,可以参见 youtube 上的这个视频。这个算法巧妙在把二维数组按行或列拆成多个一维数组,然后利用一维数组的累加和来找符合要求的数字,这里用了 lower_bound 来加快的搜索速度,也可以使用二分搜索法来替代。建立一个 TreeSet,然后开始先放个0进去,为啥要放0呢,因为要找 lower_bound(curSum - k),当 curSum 和k相等时,0就可以被返回了,这样就能更新结果了。由于对于一维数组建立了累积和,那么 sum[i,j] = sum[i] - sum[j],其中 sums[i,j] 就是目标子数组需要其和小于等于k,然后 sums[j] 是 curSum,而 sum[i] 就是要找值,当使用二分搜索法找 sum[i] 时,sum[i] 的和需要大于等于 sum[j] - k,所以也可以使用 lower_bound 来找,参见代码如下:
解法二:
class Solution {
public:
int maxSumSubmatrix(vector<vector<int>>& matrix, int k) {
if (matrix.empty() || matrix[].empty()) return ;
int m = matrix.size(), n = matrix[].size(), res = INT_MIN;
for (int i = ; i < n; ++i) {
vector<int> sum(m);
for (int j = i; j < n; ++j) {
for (int k = ; k < m; ++k) {
sum[k] += matrix[k][j];
}
int curSum = ;
set<int> st{{}};
for (auto a : sum) {
curSum += a;
auto it = st.lower_bound(curSum - k);
if (it != st.end()) res = max(res, curSum - *it);
st.insert(curSum);
}
}
}
return res;
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/363
类似题目:
Range Sum Query 2D - Immutable
Maximum Size Subarray Sum Equals k
参考资料:
https://leetcode.com/problems/max-sum-of-rectangle-no-larger-than-k/