題目：

TinyURL is a URL shortening service where you enter a URL such as https://leetcode.com/problems/design-tinyurl and it returns a short URL such as http://tinyurl.com/4e9iAk.
Design the encode and decode methods for the TinyURL service. There is no restriction on how your encode/decode algorithm should work. You just need to ensure that a URL can be encoded to a tiny URL and the tiny URL can be decoded to the original URL.

心得：

這題實務上的話我應該會用資料庫來存key，簡單易用不是嗎？
不過要小心的是不能使用流水號來當作tinyurl的參數，
這樣會有被猜出key的可能性，可能就隨機產生一個短的字串並檢查不能重複來當key

我的答案(轉base64)：

public class Codec {

    // Encodes a URL to a shortened URL
    public string encode(string longUrl) {
        return Convert.ToBase64String(Encoding.UTF8.GetBytes(longUrl));
    }

    // Decodes a shortened URL to its original URL.
    public string decode(string shortUrl) {
        return Encoding.UTF8.GetString(Convert.FromBase64String(shortUrl));
    }
}

我的答案(guid)：

public class Codec {
    
    private static Dictionary<string, string> dty = new Dictionary<string, string>();
    
    // Encodes a URL to a shortened URL
    public string encode(string longUrl) {
        string guid = Guid.NewGuid().ToString();
        dty.Add(guid, longUrl);
        return guid;
    }

    // Decodes a shortened URL to its original URL.
    public string decode(string shortUrl) {
        string url = string.Empty;
        return dty.TryGetValue(shortUrl, out url) ? url : null;
    }
}

問題：

Given two binary trees and imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not.

You need to merge them into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the NOT null node will be used as the node of new tree.

心得：

這題就是把兩個二元樹合併，我就想著最簡單遞迴來解決，看了最佳解答才發現如果為一邊null的話就可以直接用另外一邊來return更有效率，學習了。
我的答案：

public class Solution {
    public TreeNode MergeTrees(TreeNode t1, TreeNode t2) {
        if (t1 == null && t2 == null)
        {
            return null;
        }

        int t1Val = (t1 == null) ? 0 : t1.val;
        int t2Val = (t2 == null) ? 0 : t2.val;

        return new TreeNode(t1Val + t2Val)
        {
            left = MergeTrees(t1?.left, t2?.left),
            right = MergeTrees(t1?.right, t2?.right)
        };
    }
}

最佳解答：

public class Solution {
    public TreeNode MergeTrees(TreeNode t1, TreeNode t2) {
        if (t1 == null)
        {
            return t2;
        }
        if (t2 == null)
        {
            return t1;
        }

        return new TreeNode(t1.val + t2.val)
        {
            left = MergeTrees(t1.left, t2.left),
            right = MergeTrees(t1.right, t2.right)
        };
    }
}

心得

這題我本來是想用兩個for迴圈解決的

public class Solution {
    public int[] TwoSum(int[] numbers, int target) {
        for (int i = 0; i < numbers.Length; i++)
        {
            for (int j = numbers.Length - 1; j > i; j--)
            {
                if (numbers[i] + numbers[j] == target)
                {
                    return new int[] { i + 1, j + 1 };
                }
            }
        }

        throw new Exception();
    }
}

但發現會Time Limit Exceeded於是乎看了一下Solutions他們是用一個while迴圈來解答，題目要求index1<index2且不能使用重複的元素，所以這題如果用while的話宣告兩個變數分別是left與right就可以只用一個迴圈就解決。

問題

Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution and you may not use the same element twice.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

答案

public class Solution {
    public int[] TwoSum(int[] numbers, int target) {
        int left = 0;
        int right = numbers.Length - 1;
        while (numbers[left] + numbers[right] != target)
        {
            if (numbers[left] + numbers[right] > target)
            {
                right--;
            }
            else
            {
                left++;
            }
        }

        return new int[] { left + 1, right + 1 };
    }
}

心得

獲得字串並判斷字串內的字母能組成最長的回文大小為多少，例如傳入asdasdAAAvzxh則答案為9，字串為asdAvAdsa，其中v可替換成任意一位使用字母。

問題

Given a string which consists of lowercase or uppercase letters, find the length of the longest palindromes that can be built with those letters.

This is case sensitive, for example "Aa" is not considered a palindrome here.

Note:
Assume the length of given string will not exceed 1,010.

Example:

Input:
"abccccdd"

Output:
7

Explanation:
One longest palindrome that can be built is "dccaccd", whose length is 7.

答案

public class Solution {
    public int LongestPalindrome(string s) {
        int ans = 0;
        int[] lower = new int[26];
        int[] upper = new int[26];
        foreach (var item in s)
        {
            if (item > 'Z')
            {
                lower[item - 'a']++;
            }
            else
            {
                upper[item - 'A']++;
            }

        }
        for (int i = 0; i < lower.Length; i++)
        {
            ans += lower[i] / 2;
            ans += upper[i] / 2;
        }

        ans = ans * 2;
        return ans < s.Length ? ans + 1 : ans;
    }
}