我有一个这样的对象数组:聚合结果给我一个结构相同的结果如下:results=[{id:1,test:biology,candidates:[{cid:11},{cid:12},{cid:13}]},{id:2,test:chemistry,candidates:[{cid:15},{cid:16},{cid:17}]},{id:3,test:physics,candidates:[{cid:1},{cid:6},{cid:7}]}];所以我需要在数组中循环,然后为每个候选人调用一个promise函数getTotalMarksPerCandidate(它有一个Promise.all并在一
题目链接:https://leetcode-cn.com/problems/combination-sum/给定一个无重复元素的数组candidates和一个目标数target,找出candidates中所有可以使数字和为target的组合。candidates中的数字可以无限制重复被选取。说明:所有数字(包括target)都是正整数。解集不能包含重复的组合。示例1:输入:candidates=[2,3,6,7],target=7,所求解集为:[[7],[2,2,3]]示例2:输入:candidates=[2,3,5],target=8,所求解集为:[[2,2,2,2],[2,3,3],[3,
我有一个像这样的numpy数组:candidates=array([[1,0,0,0,0,1,0,0,0,1,1,1,1,1,0,1,0,1,0,1,1,0,0,1,0,1,0,0,0,1,0,0,1,0],[0,0,1,0,0,0,0,1,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,1,0,1,1,1,1,0,1,0,1,1],[1,0,1,1,1,1,0,0,1,1,0,1,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,1,1,0,0,0,0]])我不明白candidates[0]之间有什么区别:candidates[0]=array([1,0,0