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  • 339. Nested List Weight Sum

    https://leetcode.com/problems/nested-list-weight-sum/description/

    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)

    Sol:
     
     
    Because the input is nested, it is natural to think about the problem in a recursive way. We go through the list of nested integers one by one, keeping track of the current depth d. If a nested integer is an integer n, we calculate its sum as time n×d. If the nested integer is a list, we calculate the sum of this list recursively using the same process but with depth d+1.
     
     
     

    Complexity Analysis

    The algorithm takes O(N) time, where N is the total number of nested elements in the input list. For example, the list [ [[[[1]]]], 2 ] contains 4 nested lists and 2 nested integers (1and 2), so N=6.

    In terms of space, at most O(D) recursive calls are placed on the stack, where D is the maximum level of nesting in the input. For example, D=2 for the input [[1,1],2,[1,1]], and D=3 for the input [1,[4,[6]]].

    /**
     * // This is the interface that allows for creating nested lists.
     * // You should not implement it, or speculate about its implementation
     * public interface NestedInteger {
     *
     *     // @return true if this NestedInteger holds a single integer, rather than a nested list.
     *     public boolean isInteger();
     *
     *     // @return the single integer that this NestedInteger holds, if it holds a single integer
     *     // Return null if this NestedInteger holds a nested list
     *     public Integer getInteger();
     *
     *     // @return the nested list that this NestedInteger holds, if it holds a nested list
     *     // Return null if this NestedInteger holds a single integer
     *     public List<NestedInteger> getList();
     * }
     */
    public class Solution {
        public int depthSum(List<NestedInteger> nestedList) {
            
            // DFS
            return  depthSum(nestedList, 1);
            
        }
        
        public int depthSum(List<NestedInteger> list, int depth){
            int sum = 0;
            for (NestedInteger n : list){
                if (n.isInteger()){
                    sum += n.getInteger() * depth;
                } else {
                    sum += depthSum(n.getList(), depth + 1);
                }
            }
            return sum;
            
        }
        
        
    }
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  • 原文地址:https://www.cnblogs.com/prmlab/p/7279081.html
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