LeetCode 339 Nested List Weight Sum

LeetCode 339 Nested List Weight Sum

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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)

题目给出了一个不太常见的数据结构，并且给出了对应interface的实现。看到嵌套深度作为权重，第一反应是dfs。

一般dfs如果需要求深度

1. 将深度depth作为返回值，常见于min，max tree depth这类问题中。
2. 将深度作为dfs函数的形式参数，随着dfs的深入逐渐增加。

本题需要用到第二种用法。需要注意求和的递归分解。每一次都可以得到一个List或者NestedInteger。若为NestedInteger则直接返回该integer的值乘以depth；若得到List，则遍历list继续向下递归求解每一个NestedInteger的值。

/**
 * // 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 getList();
 * }
 */
public class Solution {
    public int depthSum(List nestedList) {
        if (nestedList == null) return 0;
        else {
            int sum = 0;
            for (NestedInteger x : nestedList) {
                sum += dfs(x,1);       
            }
            return sum;
        }
    }
    
    public int dfs(NestedInteger nestedInt, int depth) {
        if (nestedInt.isInteger()) {
            return nestedInt.getInteger() * depth;
        } else {
            int sum = 0;
            for (NestedInteger x : nestedInt.getList()) {
                sum += dfs(x,depth+1);
            }
            return sum;
        }
    }
    
}