标签：最大数 may pre complex this hold lex int highlight
题目链接: https://leetcode.com/problems/nested-list-weight-sum/
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)
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.
/**
* // 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) {
return depthSum(nestedList, 1);
}
public int depthSum(List<NestedInteger> nestedList, int weight) {
int sum = 0;
for (NestedInteger each : nestedList) {
if (each.isInteger())
sum += each.getInteger() * weight;
else sum += depthSum(each.getList(), weight+1);
}
return sum;
}
}
[leetcode] 339. Nested List Weight Sum
标签：最大数 may pre complex this hold lex int highlight
原文地址：http://www.cnblogs.com/apanda009/p/7948211.html
