面试笔试杂项积累-leetcode 226-230

226.226- Implement Stack using Queues-Difficulty: Easy

Invert a binary tree.

4
/   \
2     7
/ \   / \
1   3 6   9

to

4
/   \
7     2
/ \   / \
9   6 3   1

Trivia:

This problem was inspired by
this original tweet by
Max Howell:
Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so fuck off.

思路

所有左右节点都反过来

先序遍历然后反过来即可

上面那段话就不译了，，，，总之，基础都掌握不好，何谈就业

/**
* Definition for a binary tree node.
* public class TreeNode {
* public int val;
* public TreeNode left;
* public TreeNode right;
* public TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode InvertTree(TreeNode root) {
if (root == null)
return root;
printTree(root);
return root;
}
void printTree(TreeNode root)
{
TreeNode temp = root.left;
root.left = root.right;
root.right = temp;

if (root.left != null)
printTree(root.left);
if (root.right != null)
printTree(root.right);

}
}

227.227-Basic Calculator II-Difficulty: Medium

Implement a basic calculator to evaluate a simple expression string.

The expression string contains only non-negative integers,

+

,

-

,

*

,

/

operators and empty spaces

. The integer division should truncate toward zero.

You may assume that the given expression is always valid.

Some examples:

"3+2*2" = 7
" 3/2 " = 1
" 3+5 / 2 " = 5

方法一

思路

224题的加强版。没有了括号，运算符变成了+-*/

把乘除先做了，当然+，-还是使乘除的结果带上正负，把结果存到栈里，最后加在一起即可

public class Solution {
public int Calculate(string s) {
int len;
if (s == null || (len = s.Length) == 0) return 0;
Stack<int> stack = new Stack<int>();
int num = 0;
char sign = '+';
for (int i = 0; i < len; i++)
{
if (char.IsDigit(s[i]))
{
num = num * 10 + s[i] - '0';
}
if ((!char.IsDigit(s[i]) && ' ' != s[i]) || i == len - 1)
{
if (sign == '-')
{
stack.Push(-num);
}
if (sign == '+')
{
stack.Push(num);
}
if (sign == '*')
{
stack.Push(stack.Pop() * num);
}
if (sign == '/')
{
stack.Push(stack.Pop() / num);
}
sign = s[i];
num = 0;
}
}

int re = 0;
foreach (int i in stack)
{
re += i;
}
return re;
}
}

方法二

思路

又在discuss上发现了一个不需要额外数据结构辅助的方法

参考：
https://leetcode.com/discuss/42643/my-16-ms-no-stack-one-pass-short-c-solution
class Solution {
public:
int calculate(string s) {
int result = 0, cur_res = 0;
char op = '+';
for(int pos = s.find_first_not_of(' '); pos < s.size(); pos = s.find_first_not_of(' ', pos)) {
if(isdigit(s[pos])) {
int tmp = s[pos] - '0';
while(++pos < s.size() && isdigit(s[pos]))
tmp = tmp*10 + (s[pos] - '0');
switch(op) {
case '+' : cur_res += tmp; break;
case '-' : cur_res -= tmp; break;
case '*' : cur_res *= tmp; break;
case '/' : cur_res /= tmp; break;
}
}
else {
if(s[pos] == '+' || s[pos] == '-') {
result += cur_res;
cur_res = 0;
}
op = s[pos++];
}
}
return result + cur_res;
}
};

228.228-Summary Ranges-Difficulty: Easy

Given a sorted integer array without duplicates, return the summary of its ranges.

For example, given

[0,1,2,4,5,7]

, return

["0->2","4->5","7"].

思路

两个指针 start, end.  如果nums[end+1] = nums[end]+1, 就移动end指针, 否则, 插入字符串nums[start]->nums[end].

注意如果不符合情况并且start == end就保留(存入list)该数字，再进行下一个判断

public class Solution {
public IList<string> SummaryRanges(int[] nums) {
int start = 0;
int end = 0;
IList<string> fin = new List<string>();
while (end < nums.Length)
{
if (start == end && end == nums.Length - 1)
{
fin.Add(nums[end].ToString()); break;
}
if (nums[end + 1] == nums[end] + 1)
{
++end;
if (end == nums.Length - 1)
{
fin.Add(nums[start] + "->" + nums[end]);
break;
}
}
else
{
if (start == end)
{
fin.Add(nums[end].ToString());
++end; ++start;
}
else
{
fin.Add(nums[start] + "->" + nums[end]);
start = end + 1;
++end;
continue;
}
}

}
return fin;
}
}

229.229-Majority Element II-Difficulty: Easy

Given an integer array of size n, find all elements that appear more than

⌊ n/3 ⌋

times. The algorithm should run in linear time and in O(1) space.

Hint:

How many majority elements could it possibly have?

思路

169题的加强版

169题是存在重复数多余n/2,这回是多余n/3的都要，就有可能是一个数或两个数。

我们用hash表来解决，key为该数的值，value为个数，超过n/3就存入结果list

public class Solution {
public IList<int> MajorityElement(int[] nums) {
Hashtable hash = new Hashtable();
IList<int> list = new List<int>();
int fin = nums.Length / 3;
for (int i = 0; i < nums.Length; i++)
{
if (list.Contains(nums[i]))
continue;
if (hash.ContainsKey(nums[i]))
{

hash[nums[i]] = (int)hash[nums[i]] + 1;
if ((int)hash[nums[i]] > fin)
list.Add(nums[i]);
}
else
{
hash.Add(nums[i], 1);
if (1 > fin)
list.Add(nums[i]);
}
}

return list;
}
}

230.230-Kth Smallest Element in a BST-Difficulty: Medium

Given a binary search tree, write a function

kthSmallest

to find the kth smallest element in it.

Note:

You may assume k is always valid, 1 ≤ k ≤ BST's total elements.

Follow up:

What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequently? How would you optimize the kthSmallest routine?

Hint:

Try to utilize the property of a BST.
What if you could modify the BST node's structure?
The optimal runtime complexity is O(height of BST).

思路

找到在二叉搜索树中第k小的节点

看了一个hint，想到了二叉搜索树的性质，中序遍历有顺序啊，，故如此解决

/**
* Definition for a binary tree node.
* public class TreeNode {
* public int val;
* public TreeNode left;
* public TreeNode right;
* public TreeNode(int x) { val = x; }
* }
*/
public class Solution {
int count = 0;
int fin = 0;
int _k = 0;
public int KthSmallest(TreeNode root, int k)
{
_k = k;
printNode(root);
return fin;
}
public void printNode(TreeNode root)
{
if (root.left != null)
printNode(root.left);
++count;
if (count == _k)
{
fin = root.val;
return;
}
if (root.right != null)
printNode(root.right);
}
}