vector<int>ret; size_t len=A.size();longlongsum = A[0], csum = A[0];ints =0, e =0, ss =0, ee =0;//for max continuouslonglongsum0= A[0], csum0= A[0];ints0=0, e0=0, ss0=0, ee0=0;//for min continuousboolbAllNeg =true;for(inti =1; i < len; i++) {if(...
classSolution {public:intminSubArrayLen(ints, vector<int>&nums) {intlen =nums.size();if(len ==0)return0;//sums[i] : number sum from 0...ivector<int> sums(len,0); sums[0] = nums[0];for(inti =1; i < len; i++) sums[i]= sums[i -1] +nums[i];if(sums.back() < s...
classSolution {public:/** * @param nums: A list of integers * @return: A list of integers includes the index of the first number * and the index of the last number*/vector<int> subarraySumClosest(vector<int>nums) {if(nums.empty()) {return{}; }constintn =nums.size(); vector<pai...
AI检测代码解析 classSolution {public:/** * @param A an integer array * @return A list of integers includes the index of * the first number and the index of the last number*/vector<int> continuousSubarraySum(vector<int>&A) { vector<int>ret; size_t len=A.size();intcurr = A[0];...
classSolution {public:intmaxSubArray(intA[],intn) {//dp[i + 1] = max(dp[i] + A[i], A[i]);//int start = 0, end = 0;intmax = A[0];intsum = A[0];for(inti =1; i < n; i++) {if(A[i] >(sum +A[i])) ...
And here is my solution: O(n) by using Greedy strategy on this equation: sum[i..j] = sum[0..j] - sum[0..i-1]. classSolution {public:/** * @param nums: a list of integers * @return: A integer denote the sum of minimum subarray*/intminSubArray(vector<int>nums) ...