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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：
采用DFS. Base case是list中的当前item, 是integer, 乘以depth加到res中return. 若不是integer, 就对其depth+1做dfs.
Time Complexity: O(n). Space: O(n).
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 if(nestedList == null || nestedList.size() == 0){ 21 return 0; 22 } 23 int [] res = {0}; 24 for(NestedInteger item : nestedList){ 25 dfs(item, 1, res); 26 } 27 return res[0]; 28 } 29 private void dfs(NestedInteger nestInt, int depth, int [] res){ 30 if(nestInt.isInteger()){ 31 res[0] += depth * nestInt.getInteger(); 32 return; 33 } 34 for(NestedInteger nestedItem : nestInt.getList()){ 35 dfs(nestedItem, depth+1, res); 36 } 37 } 38 }
LeetCode Nested List Weight Sum
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原文地址：http://www.cnblogs.com/Dylan-Java-NYC/p/5349215.html
