题目地址:https://leetcode-cn.com/problems/nested-list-weight-sum/

题目描述

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:

Input: [[1,1],2,[1,1]]
Output: 10
Explanation: Four 1's at depth 2, one 2 at depth 1.

Example 2:

Input: [1,[4,[6]]]
Output: 27
Explanation: One 1 at depth 1, one 4 at depth 2, and one 6 at depth 3; 1 + 4*2 + 6*3 = 27.

题目大意

给定一个嵌套的整数列表,请返回该列表按深度加权后所有整数的总和。

每个元素要么是整数,要么是列表。同时,列表中元素同样也可以是整数或者是另一个列表。

解题方法

dfs

既然是个嵌套的结构,那么最简单的肯定是使用dfs去解决。函数的输入是个vector,所以对里面的每个元素进行遍历,判断是整数还是嵌套列表,如果是整数,把结果累加上整数*depth,否则dfs即可。

C++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:
* // Constructor initializes an empty nested list.
* NestedInteger();
*
* // Constructor initializes a single integer.
* NestedInteger(int value);
*
* // 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;
*
* // Set this NestedInteger to hold a single integer.
* void setInteger(int value);
*
* // Set this NestedInteger to hold a nested list and adds a nested integer to it.
* void add(const NestedInteger &ni);
*
* // 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<NestedInteger> &getList() const;
* };
*/
class Solution {
public:
int depthSum(vector<NestedInteger>& nestedList) {
return dfs(nestedList, 1);
}
int dfs(vector<NestedInteger>& nestedList, int depth) {
if (nestedList.empty())
return 0;
int res = 0;
for (auto& nest : nestedList) {
if (nest.isInteger()) {
res += nest.getInteger() * depth;
} else {
res += dfs(nest.getList(), depth + 1);
}
}
return res;
}
};

日期

2019 年 9 月 19 日 —— 举杯邀明月,对影成三人

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