Problem Difinition: Find the contiguous subarray within an array (containing at least one number) which has the largest sum. 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. More practice: If you have fig...
classSolution(object): defmaxSubArray(self, nums): """ :type nums: List[int] :rtype: int """ maxsum=nums[0] n=len(nums) foriinrange(1,n): nums[i]=max(nums[i],nums[i]+nums[i-1]) maxsum=max(maxsum,nums[i]) returnmaxsum...
此时由于题目要求数组连续,所以没法保留原sum,所以只能让sum等于从A[i]开始的新的一段数了,这一段数字形成新的candidate。 3. 如果每次得到新的candidate都和全局的max_sum进行比较,那么必然能找到最大的max sum subarray. 在循环过程中,用max_sum记录历史最大的值。从A[0]到A[n-1]一步一步地进行。 其实...
53. Maximum Subarray Problem Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum. Example: Input: [-2,1,-3,4,-1,2,1,-5,4], Output: 6 Expl......
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]; ...
or max. By keeping watch over both ends, we have the maximum product of subarrays. This holds only for integers. Given elements between [0, 1), the product could shrink. Still, using log() to convert the problem is obviously a feasible method, just know that the 0s are to be ...
class Solution { public: int maxSubArray(vector<int>& nums) { int res_left = 0; // 存最后的结果 int res_right = 0; int left = 0; // 记录当前计算和的区间 int right = -1; int res = INT_MIN; int sum = 0; for(int i=0; i<nums.size(); i++) { sum += nums[i]; /...
public int maxSubArray(int[] nums) { if(nums==null || nums.length == 0) return 0; int[] dp = new int[nums.length]; dp[0] = nums[0]; int res = dp[0]; for(int i = 1; i<nums.length;i++){ dp[i] = nums[i] + (dp[i-1]<0 ? 0: dp[i-1]); ...
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0053-maximum-subarray Time: 67 ms (90.80%), Space: 70.2 MB (29.39%) - LeetHub May 30, 2024 0056-merge-intervals Time: 27 ms (33.59%), Space: 22.6 MB (68.49%) - LeetHub May 31, 2024 0058-length-of-last-word Clean up repo Jan 21, 2024 ...