当k = 2 时, 为数组最大前缀和和最大后缀和的和 当k > 2 时, 如果数组的和为正, 需要加上 sum * (k - 2) 时间复杂度为 O(n), 空间复杂度为 O(n) 代码: C++: class Solution{public:intkConcatenationMaxSum(vector<int>&arr,intk){intn=arr.size(),pre=0,result=0,mod=1e9+7,sum=0...
classSolution{public:vector<int>maxSumOfThreeSubarrays(vector<int>&nums,int k){vector<int>ans;int sum1=0,maxSum1=0,maxSum1Idx=0;int sum2=0,maxSum12=0,maxSum12Idx1=0,maxSum12Idx2=0;int sum3=0,maxTotal=0;for(int i=k*2;i<nums.size();++i){sum1+=nums[i-k*2];sum2+=n...
publicintmaxSum1(int[]arry,int k){int size=arry.length;if(size<k){return-1;}//初始化第一个窗口值的总和int maxSum=0;for(int i=0;i<k;i++){maxSum=maxSum+arry[i];}int sum=maxSum;for(int i=k;i<size;i++){sum=sum+arry[i]-arry[i-k];maxSum=Math.max(maxSum,sum);}retur...
数值的初始化,一开始先对sum进行一个赋值,但有可能初始情况并不满足要求,最终导致结果被提前内定从而导致错误,但是用flag==1有点丑其实 整体而言用了太多if else的判断 class Solution: def maxSum(self, nums,m,k): lens = len(nums) #print(lens,k) if lens < k: #print('-1') return 0 counter ...
不过这个dp状态矩阵很特殊，特殊到我们用一个max_sum来储存就行并不需要一个向量：举例：{−1, 1,...
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]; i < n; ++i) {if...
int maxLevelSum(TreeNode*root) { int res = 1, mx = root->val, level = 0; queue<TreeNode*> q{{root}}; while (!q.empty()) { ++level; int sum = 0; for (inti=q.size(); i > 0;--i) {TreeNode* t = q.front(); q.pop(); ...
这里是python代码实现:pythonclass Solution: def maximizeSum(self, nums: List[int], k: int) -> int: m = max(nums) return (2*m + k - 1) * k // 2解答思路:1. 首先找到数组中的最大值m。2. 然后最大的得分公式为:m + (m+1) + (m+2) + ... + (m+k-1) = (m...
321. 拼接最大数ctrl c,v 很快啊 看
[int], 输入的整数数组:return: int, 最大子数组的和"""if not nums:return 0n = len(nums)dp = nums[0]max_sum = dpfor i in range(1, n):dp = max(nums[i], dp + nums[i])if dp > max_sum:max_sum = dpreturn max_sum# 示例调用print(maxSubArray([-2,1,-3,4,-1,2,1,-5...