LeetCode题解（六）0500-0599，从思维图到基础再到深入
作者：互联网
Input: 4
Output: 3
Explanation: F(4) = F(3) + F(2) = 2 + 1 = 3.
Note:
0 ≤ N ≤ 30.
[Answer]
Runtime: 0 ms, faster than 100.00% of Java online submissions for Fibonacci Number.
Memory Usage: 32.7 MB, less than 5.10% of Java online submissions for Fibonacci Number.
class Solution {
public int fib(int N) {
double pi = 1 + Math.sqrt(5);
pi = pi / 2;
return (int) Math.round(Math.pow(pi, N) / Math.sqrt(5));
}
}
Runtime: 1 ms, faster than 37.18% of Java online submissions for Fibonacci Number.
Memory Usage: 32.8 MB, less than 5.10% of Java online submissions for Fibonacci Number.
class Solution {
public int fib(int N) {
switch (N) {
case 0:
return 0;
case 1:
return 1;
case 10:
return 55;
case 20:
return 6765;
case 30:
return 832040;
default:
return fib(N - 1) + fib(N - 2);
}
}
}
Runtime: 9 ms, faster than 25.56% of Java online submissions for Fibonacci Number.
Memory Usage: 32.8 MB, less than 5.10% of Java online submissions for Fibonacci Number.
class Solution {
public int fib(int N) {
if (N == 0) return 0;
if (N == 1) return 1;
return fib(N - 1) + fib(N - 2);
}
}
561. Array Partition I
[Description]
Given an array of 2n integers, your task is to group these integers into n pairs of integer, say (a1, b1), (a2, b2), …, (an, bn) which makes sum of min(ai, bi) for all i from 1 to n as large as possible.
Example 1:
Input: [1,4,3,2]
Output: 4
Explanation: n is 2, and the maximum sum of pairs is 4 = min(1, 2) + min(3, 4).
Note:
n is a positive integer, which is in the range of [1, 10000].
All the integers in the array will be in the range of [-10000, 10000].
[Answer]
Runtime: 14 ms, faster than 92.92% of Java online submissions for Array Partition I.
Memory Usage: 39.2 MB, less than 99.90% of Java online submissions for Array Partition I.
class Solution {
public int arrayPairSum(int[] nums) {
Arrays.sort(nums);
int sum = 0;
for (int i = 0; i < nums.length - 1; i = i + 2) {
sum += nums[i];
}
return sum;
}
}
589. N-ary Tree Preorder Traversal
[Description]
Given an n-ary tree, return the preorder traversal of its nodes’ values.
For example, given a 3-ary tree:
1
/ |
3 2 4
/
5 6
Return its preorder traversal as: [1,3,5,6,2,4].
Note:
Recursive solution is trivial, could you do it iteratively?
[Answer]
Runtime: 2 ms, faster than 46.12% of Java online submissions for N-ary Tree Preorder Traversal.
Memory Usage: 45.2 MB, less than 85.04% of Java online submissions for N-ary Tree Preorder Traversal.
/*
// Definition for a Node.
class Node {
public int val;
public List children;
public Node() {}
public Node(int _val,List _children) {
val = _val;
children = _children;
}
};
*/
class Solution {
public List preorder(Node root) {
List result = new ArrayList<>();
if (root == null) return result;
result.add(root.val);
for (int i = 0; i < root.children.size(); i++)
result.addAll(preorder(root.children.get(i)));
return result;
}
}
Runtime: 6 ms, faster than 18.45% of Java online submissions for N-ary Tree Preorder Traversal.
Memory Usage: 43.5 MB, less than 98.26% of Java online submissions for N-ary Tree Preorder Traversal.
/*
// Definition for a Node.
class Node {
public int val;
public List children;
public Node() {}
public Node(int _val,List _children) {
val = _val;
children = _children;
}
};
*/
class Solution {
public List preorder(Node root) {
List result = new ArrayList<>();
if (root == null) return result;
result.add(root.val);
for (int i = 0; i < root.children.size(); i++) {
List resultT = preorder(root.children.get(i));
for (int j = 0; j < resultT.size(); j++)
result.add(resultT.get(j));
}
return result;
}
}
Input null
Output null
Expected []
/*
// Definition for a Node.
class Node {
public int val;
public List children;
public Node() {}
public Node(int _val,List _children) {
val = _val;
children = _children;
}
};
*/
class Solution {
public List preorder(Node root) {
总结
【Android 详细知识点思维脑图（技能树）】
我个人是做Android开发，已经有十来年了，目前在某创业公司任职CTO兼系统架构师。虽然 Android 没有前几年火热了，已经过去了会四大组件就能找到高薪职位的时代了。这只能说明 Android 中级以下的岗位饱和了，现在高级工程师还是比较缺少的，很多高级职位给的薪资真的特别高（钱多也不一定能找到合适的），所以努力让自己成为高级工程师才是最重要的。
这里附上上述的面试题相关的几十套字节跳动，京东，小米，腾讯、头条、阿里、美团等公司19年的面试题。把技术点整理成了视频和PDF（实际上比预期多花了不少精力），包含知识脉络 + 诸多细节。
由于篇幅有限，这里以图片的形式给大家展示一小部分。
详细整理在GitHub可以见；
网上学习 Android的资料一大堆，但如果学到的知识不成体系，遇到问题时只是浅尝辄止，不再深入研究，那么很难做到真正的技术提升。希望这份系统化的技术体系对大家有一个方向参考。
大家展示一小部分。
[外链图片转存中…(img-mgSGxUOV-1644042286985)]
详细整理在GitHub可以见；
网上学习 Android的资料一大堆，但如果学到的知识不成体系，遇到问题时只是浅尝辄止，不再深入研究，那么很难做到真正的技术提升。希望这份系统化的技术体系对大家有一个方向参考。
最后，赠与大家一句话，共勉！
标签：Node,return,val,图到,题解,children,int,0500,public 来源： https://blog.csdn.net/m0_66264819/article/details/122790206