原题链接在这里：https://leetcode.com/problems/nested-list-weight-sum/
题目：
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)
题解：
For dfs state, it needs current nested list and current depth.
For each NestedInteger ni in the list, if it is integer, add its value * depth to res. Otherwise, continue DFS with it and depth+1.
Time Complexity: O(n). n 是指全部叶子的数目加上dfs走过层数的总数. [[[[[5]]]],[[3]], 1], 3个叶子, dfs一共走了6层. 所以用了 3 + 6 = 9 的时间.
Space: O(D). D 是recursive call用的stack的最大数目, 即是最深的层数, 上面例子最深走过4层, 这里D = 4.
AC Java:
/**
* // 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) {
return dfs(nestedList, 1);
}
private int dfs(List<NestedInteger> nestedList, int depth){
int sum = 0;
for(NestedInteger item : nestedList){
if(item.isInteger()){
sum += item.getInteger()*depth;
}else{
sum += dfs(item.getList(), depth+1);
}
}
return sum;
}
}