publicclassSolution {publicArrayList<Integer> subArraySum (int[] nums) { ArrayList<Integer> ans =newArrayList<>();if(nums ==null|| nums.length == 0) {return0; } HashMap<Integer, Integer> map =newHashMap<>(); map.put(0, -1);intsum = 0;for(inti = 0; i < nums.length; i++)...
and the index of the last number"""defsubarraySum(self, nums): map= {0:-1} sum=0foriinxrange(len(nums)): sum+=nums[i]ifmap.has_key(sum):return[map[sum]+1,i] map[sum]=ireturn
class Solution { public: int subarraySum(vector<int>& nums, int k) { unordered_map<int, int> m{{0,1}}; int count=0; int sum=0; for(int num:nums){ sum+=num; count+=m[sum-k]; m[sum]++; } return count; } }; 523. Continuous Subarray Sum 这题问的是是否有subarray的sum...
Leetcode力扣 209 | 长度最小的子数组 Minimum Size Subarray Sum 2020年11月05日 07:455743浏览·30点赞·11评论 爱学习的饲养员 粉丝:7.0万文章:46 关注 视频讲解 622:17 Leetcode力扣 1-300题视频讲解合集|手画图解版+代码【持续更新ing】 84.4万798 ...
class Solution { public: int subarraySum(vector<int>& nums, int k) { int n = nums.size(), sum = 0, res = 0; vector<int> pre(n); unordered_map<int, int> mp; mp[0] = 1; for(int i = 0; i < n; ++i) { sum += nums[i]; pre[i] = sum; } for(int i = 0; i...
Subarray Sum Given an integer array, find a subarray where the sum of numbers iszero. Your code should return the index of the first number and the index of the last number. Notice There is at least one subarray that it's sum equals to zero. ...
Sub-array from 0 to 3 and sum is: 10 Sample Solution: Scala Code: objectScala_Array{deffind_min_subarray_sum(nums:Array[Int],k:Int):Array[Int]={varsub_arr_sum=0;varmin_sub_arr=Integer.MAX_VALUE;varlast=0;varresult=newArray[Int](3)for(i<-0tonums.length-1){sub_arr_sum+=nums...
longlongsum=0;deque<pair<;intret=INT_MAX;for(inti=0;i<n;++i){while(!q.empty()&&q.back().second>=sum)q.pop_back();q.push_back(make_pair(i,sum));sum+=A[i];while(!q.empty()&&q.front().second<=sum-k){ret=min(ret,i+1-q.front().first);q.pop_front();}}returnret=...
leftMax = max(leftMax, sumNum) rightMax = sumNum = 0 for i in range(m + 1, r + 1): # 从中间向右遍历 sumNum += nums[i] rightMax = max(rightMax, sumNum) leftAns = self.maxSubArrayHelper(nums, l, m - 1) rightAns = self.maxSubArrayHelper(nums, m + 1, r) ...
*/vector<int>subarraySum(vector<int>&nums){// write your code hereunordered_map<int,int>umap;intcur_sum=0;vector<int>result;for(inti=0;i<nums.size();i++){cur_sum+=nums[i];if(cur_sum==0){result.push_back(0);result.push_back(i);break;}if(umap.find(cur_sum)==umap.end())...