We define the pivot index as the index where the sum of the numbers to the left of the index is equal to the sum of the numbers to the right of the index. If no such index exists, we should return -1. If there are multiple pivot indexes, you should return the left-most pivot in...
Calculate prefix sumBof listA. B[j] - B[i]represents the sum of subarrayA[i] ~ A[j-1] Dequedwill keep indexes of increasingB[i]. For every B[i], we will compareB[i] - B[d[0]]withK. publicintshortestSubarray(int[] A,intK) {intN = A.length, res = N + 1;int[] B ...
在第一回 a ^ b 之后, b 的作用就消失了. 接下去应该给parameter重新命名. sum = a ^ b; // sum without adding carries nextCarry = (a & b) << 1; 用其他variable name 取代 a, b 会更好理解一点. Bit Operation Steps: a & b: 每bit可能出现的余数 a ^ b: 每bit在此次操作可能留下的...
sliding window maximum: this problem uses deque, but the idea is the same: keep a decreasing sequence of indexes calculator series evaluate reverse polish notation: stack basic calculator: solved with two stacks: one for operands one for numbers. Can be solved with just one stack. Use number ...
解释: 子数组 [-1, 2] 和等于 1,且长度最长。 进阶: 你能使时间复杂度在 O(n) 内完成此题吗? 来源:力扣(LeetCode) 链接:力扣 classSolution{public:intmaxSubArrayLen(vector<int>&nums,intk){unordered_map<int,int>hash;hash[0]=-1;intans=0;intsum=0;for(inti=0;i<nums.size();i++){sum...
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- We need to store the relationship between two sets of data in order to compute the required result. What is the slicing for this array if mid = 1?[3,9,20,15,7]preorder[1:mid+1] [9]- Slicing is NOT inclusive on the right hand side! [1:2] = [1,1] in the indexes. ...
But unfortunately, due to the data error, one of the numbers in the set got duplicated to another number in the set, which results in repetition of one number and loss of another number.Given an array nums representing the data status of this set after the error. Your task is to ...
We define the pivot index as the index where the sum of the numbers to the left of the index is equal to the sum of the numbers to the right of the index. If no such index exists, we should return -1. If there are multiple pivot indexes, you should return the left-most pivot in...
// insert all first indexes of arrays into the heap for (int i = 0; i < k; i++) { Node *tmp = new Node(kArrays[i][0], i, 0); minheap.push(tmp); } vector<int> ans; // step 2 : further processing while (minheap.size() > 0) { Node *temp = minheap.top(); ans...