Easy.
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 + 42 + 63 = 27)
做这个题是做到Nested List Weight Sum ii的时候看到有人在讨论，没想到又暴露了自己基础知识的问题。
Screen Shot 2017-10-19 at 5.13.31 PM.png
这种写法问题出在int传递参数的时候，是将传入的参数的具体值（基本数据类型）。所以形参接收并复制其具体的int值，在方法内部对int值进行了更改，直接反映到了其方法内部的变量内容上，对方法外部的实参是没有影响的。比如说这里，我们遇到[1,1]的时候，进入helper函数，计算出res = 2, 但接着遍历到2的时候，这个res又会回到本身的值res = 0, 所以res最后还是等于0.
下面这篇文章很好地解释了Java的传参方式：
Java传参方式
public class Solution {
public int depthSum(List<NestedInteger> nestedList) {
// Write your code here
int depth = 1;
int res = 0;
helper(nestedList, depth, res);
return res;
}
private void helper(List<NestedInteger> nestedList, int depth, int res){
for (NestedInteger ni : nestedList){
if (!ni.isInteger()){
helper(ni.getList(), depth + 1, res);
} else {
res += ni.getInteger() * depth;
}
}
}
}
