Runtime:40 ms, faster than74.57% of Python3 online submissions for Running Sum of 1d Array. Memory Usage:14 MB, less than100.00% of Python3 online submissions for Running Sum of 1d Array. 【解法三】 在原来的array里改的
【leetcode】1480. Running Sum of 1d Array 题目如下: Given an arraynums. We define a running sum of an array asrunningSum[i] = sum(nums[0]…nums[i]). Return the running sum ofnums. Example 1: Input: nums = [1,2,3,4] Output: [1,3,6,10] Explanation: Running sum is obtained...
class Solution: def runningSum(self, nums: List[int]) -> List[int]: for i in range(1,len(nums)): nums[i] += nums[i-1] return nums 1. 2. 3. 4. 5.
首先,由于直接使用从 java 传过来的 array 非常麻烦,还得调 java 的方法,所以将从 java 中拿到的 array 转为 c++ 的数据格式: intlen1 = env -> GetArrayLength(lifeList);autodim = (jintArray)env->GetObjectArrayElement(lifeList,0);intlen2 = env -> GetArrayLength(dim);int**board; board = n...
*https://leetcode.com/problems/running-sum-of-1d-array/* * Given an array nums. We define a running sum of an array as runningSum[i] = sum(nums[0]…nums[i]). Return the running sum of nums. Example 1: Input: nums = [1,2,3,4] ...
1480. Running Sum of 1d Array solution #1: code: 参考 1.leetcode_1480. Running Sum of 1d Array; 完 各美其美,美美与共,不和他人作比较,不对他人有期待,不批判他人,不钻牛角尖。 心正意诚,做自己该做的事情,做自己喜欢做的事情,安静做一枚有思想的技术媛。
LeetCode #1480. Running Sum of 1d Array 题目1480. Running Sum of 1d Array解题方法简单的动态规划题。 时间复杂度:O(n) 空间复杂度:O(1)代码class Solution: def runningSum(self, nums: List[int]) -> List[int]: for i in range(1, len(nums)): nums[i] += nums[i-1] return nums...
两种方法在leetcode都能AC。一般来说如果C++不希望修改传入的数组,那么函数签名 vector<int> runningSum(const vector<int>& nums) 会更合适,但实际生产环境中,这样的函数设计,不改变传来的函数参数是常态。否则,相当于这个函数包含有 side-effect 。如果被老师或者面试官问到这种问题,或许讨论这方面要比给出答案...
链接:https://leetcode.cn/problems/running-sum-of-1d-array 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。 题解1: 查看代码 classSolution{/** *@paramInteger[] $nums *@returnInteger[] */functionrunningSum($nums){$new_list= [];if(empty($nums))return$new_list;$sum=0...
1480. Running Sum of 1d Array Given an arraynums. We define a running sum of an array asrunningSum[i] = sum(nums[0]…nums[i]). Return the running sum ofnums. Example 1: Input: nums = [1,2,3,4] Output: [1,3,6,10]