• LeetCode 339. Nested List Weight Sum


    原题链接在这里:https://leetcode.com/problems/nested-list-weight-sum/

    题目:

    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)

    题解:

    For dfs state, it needs current nested list and current depth.

    For each NestedInteger ni in the list, if it is integer, add its value * depth to res. Otherwise, continue DFS with it and depth+1.

    Time Complexity: O(n). n 是指全部叶子的数目加上dfs走过层数的总数. [[[[[5]]]],[[3]], 1], 3个叶子, dfs一共走了6层. 所以用了 3 + 6 = 9 的时间. 

    Space: O(D). D 是recursive call用的stack的最大数目, 即是最深的层数, 上面例子最深走过4层, 这里D = 4.

    AC Java:

     1 /**
     2  * // This is the interface that allows for creating nested lists.
     3  * // You should not implement it, or speculate about its implementation
     4  * public interface NestedInteger {
     5  *
     6  *     // @return true if this NestedInteger holds a single integer, rather than a nested list.
     7  *     public boolean isInteger();
     8  *
     9  *     // @return the single integer that this NestedInteger holds, if it holds a single integer
    10  *     // Return null if this NestedInteger holds a nested list
    11  *     public Integer getInteger();
    12  *
    13  *     // @return the nested list that this NestedInteger holds, if it holds a nested list
    14  *     // Return null if this NestedInteger holds a single integer
    15  *     public List<NestedInteger> getList();
    16  * }
    17  */
    18 public class Solution {
    19     public int depthSum(List<NestedInteger> nestedList) {
    20         return dfs(nestedList, 1);
    21     }
    22     private int dfs(List<NestedInteger> nestedList, int depth){
    23         int sum = 0;
    24         for(NestedInteger item : nestedList){
    25             if(item.isInteger()){
    26                 sum += item.getInteger()*depth;
    27             }else{
    28                 sum += dfs(item.getList(), depth+1);
    29             }
    30         }
    31         return sum;
    32     }
    33 }

    类似Employee ImportanceNested List Weight Sum II.

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  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/5349215.html
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