https://leetcode.com/problems/nested-list-weight-sum/description/
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)
Complexity Analysis
The algorithm takes O(N) time, where N is the total number of nested elements in the input list. For example, the list [ [[[[1]]]], 2 ]
contains 4 nested lists and 2 nested integers (1and 2), so N=6.
In terms of space, at most O(D) recursive calls are placed on the stack, where D is the maximum level of nesting in the input. For example, D=2 for the input [[1,1],2,[1,1]]
, and D=3 for the input [1,[4,[6]]]
.
/**
* // This is the interface that allows for creating nested lists.
* // You should not implement it, or speculate about its implementation
* public interface NestedInteger {
*
* // @return true if this NestedInteger holds a single integer, rather than a nested list.
* public boolean isInteger();
*
* // @return the single integer that this NestedInteger holds, if it holds a single integer
* // Return null if this NestedInteger holds a nested list
* public Integer getInteger();
*
* // @return the nested list that this NestedInteger holds, if it holds a nested list
* // Return null if this NestedInteger holds a single integer
* public List<NestedInteger> getList();
* }
*/
public class Solution {
public int depthSum(List<NestedInteger> nestedList) { // DFS
return depthSum(nestedList, 1); } public int depthSum(List<NestedInteger> list, int depth){
int sum = 0;
for (NestedInteger n : list){
if (n.isInteger()){
sum += n.getInteger() * depth;
} else {
sum += depthSum(n.getList(), depth + 1);
}
}
return sum; } }