Given a nested list of integers, return the sum of all integers in the list weighted by their depth.
Each element is either an integer, or a list -- whose elements may also be integers or other lists.
Example 1:
Given the list [[1,1],2,[1,1]], return 10. (four 1's at depth 2, one 2 at depth 1)
Example 2:
Given the list [1,[4,[6]]], return 27. (one 1 at depth 1, one 4 at depth 2, and one 6 at depth 3; 1 + 4*2 + 6*3 = 27)
This question is a simplified version of DFS
class Solution {
public:
int DFS(vector<NestedInteger>& nestedList, int depth)
{
int sum = 0;
for(auto val: nestedList)
{
if(val.isInteger())
sum += val.getInteger()*depth;
else sum += DFS(val.getList(), depth+1);
}
return sum;
}
int depthSum(vector<NestedInteger>& nestedList) {
return DFS(nestedList, 1);
}
};
本文介绍了一种通过递归方法解决带权重的嵌套整数列表求和问题,详细解释了算法实现,并通过两个例子展示了如何计算不同深度整数的加权和。
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