本题和560. Subarray Sum Equals K非常类似,数组中有负数并不影响。 classSolution {public:intmaxSumSubmatrix(vector<vector<int>>& matrix,intk) {intm=matrix.size(), n=m==0?0:matrix[0].size();if(m==0|| n==0)return0;intres=INT_MIN;for(intr1=0;r1<m;++r1){ vector<int> compress(...
DescriptionGiven 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 = …
classSolution {public: vector<int> maxSumOfThreeSubarrays(vector<int>& nums,intk) {intn = nums.size(), mx =INT_MIN; vector<int> sums{0}, res, left(n,0), right(n, n -k);for(intnum : nums) sums.push_back(sums.back() +num);for(inti = k, total = sums[k] - sums[0];...
'failed' p from Failed where year(fail_date)=2019 order by 1 ) select p period_state,min(d) start_date, max(d) end_date from ( select *, row_number() over() rk1, rank() over(partition by p order by d) rk2 from t1 order by d ) ...
maxsum = nums[i] + nums[-1] + nums[-2] #从i开始最大的三个值 if minsum >= target: #如果最小的三个值比target还大,后面的都比它大 if abs(minsum-target) >= abs(ans-target): #如果他们的距离比前面的小,就直接返回 return ans ...
public int minSubArrayLen(int s, int[] nums) { if (nums.length == 0) return 0; int minLen = Integer.MAX_VALUE, start = 0, end = 0; int sum = 0; while (end < nums.length) { while (sum < s && end < nums.length) { ...
minSum=(~(unsignedint))>>; my_grid=grid; rowMax=grid.size(); colMax=grid[].size(); tra(,,); returnminSum; } }; 解法2:DP(还是不熟练,不太熟练递推dp和递归dp的区别,参考文章) dp[100][100];该dp数组记录的是每个位置上的最优解,即到达这一点的路径最小值。假设我们要求以grid[i][j...
importjava.util.TreeSet;publicclassSolution{publicintmaxSumSubmatrix(int[][]matrix,intk){if(matrix==null||matrix.length==0||matrix[0].length==0){return0;}intmax=Integer.MIN_VALUE;intm=matrix.length;intn=matrix[0].length;for(inti=0;i<n;i++){int[]sum=newint[m];for(intj=i;j<n...
LeetCode第一题即为Two Sum. 求数组中和为目标值的两个数的索引下标。 题目描述 Given an array of integersnumsand an integertarget, returnindices of the two numbers such that they add up totarget. You may assume that each input would haveexactly one solution, and you may not use thesameeleme...
ans=max(ans,sum_val+x-min_s[x+k]) ifx-kinmin_s: ans=max(ans,sum_val+x-min_s[x-k]) ifxnotinmin_sorsum_valmin_s[x]=sum_val sum_val+=x ifans==float('-inf'): return0 returnans nums=[-1,3,2,4,5] k=3 print(maximum_subarray_sum(nums,k))...