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:

1. Try to utilize the property of a BST.
2. What if you could modify the BST node's structure?
3. 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);    }}