public int crossSum(int[] nums, int left, int right, int p) { if (left == right) return nums[left]; int leftSubsum = Integer.MIN_VALUE; int currSum = 0; for(int i = p; i > left - 1; --i) { currSum += nums[i]; leftSubsum = Math.max(leftSubsum, currSum); } in...
The maximum sum in the first I elements is either the maximum sum in the first i - 1 elements (which we'll call MaxSoFar), or it is that of a subvector that ends in position i (which we'll call MaxEndingHere). MaxEndingHere is either A[i] plus the previous MaxEndingHere, or ...
int maxSubArray(int A[], int n) { int ret=maxsub(A,0,n-1); return ret; } protected: int maxsub(int A[], int start, int end) { int max_left=INT_MIN,max_mid=INT_MIN,max_right=INT_MIN; if(start==end) return A[start]; if(start+1==end) { int a=max(A[start],A[end...
0695 Max Area of Island Go 62.7% Medium 0696 Count Binary Substrings 56.0% Easy 0697 Degree of an Array Go 53.8% Easy 0698 Partition to K Equal Sum Subsets 44.9% Medium 0699 Falling Squares Go 41.8% Hard 0700 Search in a Binary Search Tree 73.1% Easy 0701 Insert into a Binary...
【题目】Given a non-empty array containing only positive integers, find if the array can be partitioned into two subsets such that the sum of elements in both subsets is equal. Note: Each of the array element will not exceed 100. The array size will not exceed 200. Example 1: ...
链接:https://leetcode-cn.com/problems/split-a-string-into-the-max-number-of-unique-substrings 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。 思路是backtracking。我们需要一个hashset来判断是否存在重复的子串,回溯函数的base case是指针遍历到 input 字符串尾部,就结算 hashset 里面...
publicintmaxSubArray(int[] A) { intn = A.length; inti; intmaxSum = A[0]; intthisSum =0; for(i =0; i < n; i++) { thisSum += A[i]; if(thisSum > maxSum) maxSum = thisSum; if(thisSum <0) thisSum =0; }
For example, given the array[−2,1,−3,4,−1,2,1,−5,4], the contiguous subarray[4,−1,2,1]has the largest sum =6. 题目大意 求数组的最大子数组的和。 解题思路 动态规划问题。已知了前k个元素的最大子序列和为maxSub(已经被记录下来了),以及一个暂时和sum。假设加入了第k+1这...
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(nums,0,n-1)def__max_sub_array(self,nums,left,right):ifleft==right:returnnums[left]mid=left+(right-left)//2returnmax(self.__max_sub_array(nums,left,mid),self.__max_sub_array(nums,mid+1,right),self.__max_cross_array(nums,left,mid,right))def__max_cross_array(self,nums,left,...