原题
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 + 42 + 63 = 27.
Accepted
50,412
Submissions
75,208
解法
BFS. 初始化depth和ans, 遍历nestedList, 每次遍历将加权的和累加至ans里, 然后递增depth.
Time: O(n), n是nestList的数字个数.
Space: O(n)
代码
# """
# This is the interface that allows for creating nested lists.
# You should not implement it, or speculate about its implementation
# """
#class NestedInteger(object):
# 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(object):
def depthSum(self, nestedList):
"""
:type nestedList: List[NestedInteger]
:rtype: int
"""
depth, ans = 1, 0
while nestedList:
ans += depth*sum([i.getInteger() for i in nestedList if i.isInteger()])
newList = []
for i in nestedList:
if not i.isInteger():
newList += i.getList()
nestedList = newList
depth += 1
return ans
本文介绍了一种针对嵌套整数列表的深度加权求和算法,通过BFS遍历方式,计算列表中所有整数与其深度的乘积之和。示例包括[[1,1],2,[1,1]]和[1,[4,[6]]]的求和过程及结果。
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