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LeetCode: 339. 嵌套列表权重和

1、题目描述

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。

示例 1:

示例 1:

输入: [[1,1],2,[1,1]]
输出: 10 
解释: 因为列表中有四个深度为 2 的 1 ,和一个深度为 1 的 2。


示例 2:

输入: [1,[4,[6]]]
输出: 27 
解释: 一个深度为 1 的 1,一个深度为 2 的 4,一个深度为 3 的 6。所以,1 + 4*2 + 6*3 = 27。

2、解题思路

DFS深度优先搜索

# """
# This is the interface that allows for creating nested lists.
# You should not implement it, or speculate about its implementation
# """
#class NestedInteger:
#    def __init__(self, value=None):
#        """
#        If value is not specified, initializes an empty list.
#        Otherwise initializes a single integer equal to value.
#        """
#
#    def isInteger(self):
#        """
#        @return True if this NestedInteger holds a single integer, rather than a nested list.
#        :rtype bool
#        """
#
#    def add(self, elem):
#        """
#        Set this NestedInteger to hold a nested list and adds a nested integer elem to it.
#        :rtype void
#        """
#
#    def setInteger(self, value):
#        """
#        Set this NestedInteger to hold a single integer equal to value.
#        :rtype void
#        """
#
#    def getInteger(self):
#        """
#        @return the single integer that this NestedInteger holds, if it holds a single integer
#        Return None if this NestedInteger holds a nested list
#        :rtype int
#        """
#
#    def getList(self):
#        """
#        @return the nested list that this NestedInteger holds, if it holds a nested list
#        Return None if this NestedInteger holds a single integer
#        :rtype List[NestedInteger]
#        """

class Solution:
    def depthSum(self, nestedList: List[NestedInteger]) -> int:
        return self.dfs(nestedList,1)

    def dfs(self,nestedList: List[NestedInteger],level):
        temp = 0
        for i in nestedList:
            if i.isInteger():
                temp += level * i.getInteger()
            else:
                temp += self.dfs(i.getList(),level+1)

        return temp