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 + 42 + 63 = 27)
Solution. This can be easily solved by recursion. For each list, if is a number, add to ans[]. If not, call recursion on it.
Java solution:
public int weightedSum(List list) {
return weightedSumHelper(list, 1);
}
public int weightedSumHelper(List<Object> list, int depth) {
int ans = 0;
for (Object l : list) {
if (l instanceof List) {
ans += weightedSumHelper((List)l, depth + 1);
}
else {
int num = (Integer)l;
ans += depth * num;
}
}
return ans;
}
Python solution:
def weightedSum(self, list):
ans = [0]
self.weightedSumHelper(l, 1, ans)
return ans[0]
def weightedSumHelper(self, l, depth, ans):
for ele in l:
if isinstance(ele, list):
self.weightedSumHelper(ele, depth + 1, ans)
else:
ans[0] += ele * depth