[leetcode] 339. Nested List Weight Sum 解题报告

题目链接: https://leetcode.com/problems/nested-list-weight-sum/

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)

思路: 一个比较简单的DFS, 因为数组中的每一个元素都可能包含另外一个数组, 因此这是一个多层嵌套的关系, 但是嵌套的层数需要有个记录, 所以我们只需要另外写一个函数, 记录当前是第几层即可.

代码如下:

/**
 * // This is the interface that allows for creating nested lists.
 * // You should not implement it, or speculate about its implementation
 * class NestedInteger {
 *   public:
 *     // Return true if this NestedInteger holds a single integer, rather than a nested list.
 *     bool isInteger() const;
 *
 *     // Return the single integer that this NestedInteger holds, if it holds a single integer
 *     // The result is undefined if this NestedInteger holds a nested list
 *     int getInteger() const;
 *
 *     // Return the nested list that this NestedInteger holds, if it holds a nested list
 *     // The result is undefined if this NestedInteger holds a single integer
 *     const vector &getList() const;
 * };
 */
class Solution {
public:
    int DFS(vector& 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& nestedList) {
        return DFS(nestedList, 1);
    }
};