Leetcode

9. Palindrome Number Hardness: \(\color{green}\textsf{Easy}\) Ralated Topics: Math 一、題目 Given an integer x, return true if x is palindrome number. An integer is a palindrome when it reads the same backward as forward. For example, 121 is a palindrome while 123 is not. Example 1: Input: x = 121 Output: true Explanation: 121 reads as 121 from left to right and from right to left. Example 2: Input: x = -121 Output: false Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome. Example 3: ...

8. String to Integer (atoi) Hardness: \(\color{orange}\textsf{Medium}\) Ralated Topics: String 一、題目 Implement the myAtoi(string s) function, which converts a string to a 32-bit signed integer (similar to C/C++’s atoi function). The algorithm for myAtoi(string s) is as follows: Read in and ignore any leading whitespace. Check if the next character (if not already at the end of the string) is '-' or '+'. Read this character in if it is either. This determines if the final result is negative or positive respectively. Assume the result is positive if neither is present. Read in next the characters until the next non-digit character or the end of the input is reached. The rest of the string is ignored. Convert these digits into an integer (i.e. "123" -> 123, "0032" -> 32). If no digits were read, then the integer is 0. Change the sign as necessary (from step 2). If the integer is out of the 32-bit signed integer range [-2^31, 2^31-1], then clamp the integer so that it remains in the range. Specifically, integers less than -2^31 should be clamped to -2^31, and integer greater than 2^31-1 should be clamped to 2^31-1. Return the integer as the final result. Note: ...

7. Reverse Integer Hardness: \(\color{orange}\textsf{Medium}\) Ralated Topics: Math 一、題目 Given a signed 32-bit integer x, return x with its digits reversed. If reversing x causes the value to go outside the signed 32-bit integer range [-2^31, 2^31-1], then return 0. Assume the environment does not allow you to store 64-bit integers (signed or unsigned). Example 1: Input: x = 123 Output: 321 Example 2: Input: x = -123 Output: -321 Example 3: ...

6. Zigzag Conversion Hardness: \(\color{orange}\textsf{Medium}\) Ralated Topics: String 一、題目 The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to dispaly this pattern in a fixed font for better legibility) \( \quad\texttt{P A H N}\\ \quad\texttt{APLSIIG}\\ \quad\texttt{Y I R}\\ \) And then read line by line: "PAHNAPLSIIGYIR" Write the code that will take a string and make this conversion given a number of rows: string convert(string s, int numRows ...

5. Longest Substring Without Repeating Characters Hardness: \(\color{orange}\textsf{Medium}\) Ralated Topics: String、Dynamic Programming 一、題目 Given a string s, return the longest palindromic substring in s. A string is called a palindrome string if the reverse of that string is the same of the original string. Example 1: Input: s = “babad” Output: “bab” Explanation: “aba” is also a valid answer. Example 2: Input: s = “cbbd” Output: “bb” Constraints: 1 <= s.length <= 1000 s consists of only digits and English letters. 二、分析 注意 palindrome string 的特性： 當長度為 1 時，必為 palindrome string 當長度為 2 時，兩個字元必須相同才為 palindrome string 當長度 >2 時，palindrome string 必須滿足 最左邊的字元等於最右邊的字元，即 s[left] == s[right] 除去最左邊的字元跟最右邊的字元，必須為 palindrome string， 即 s.substr(left+1, len-2) 為 palindromic。 三、解題 1. Dynamic Prograimming Time complexity: \(O(n^2)\) Space complexity: \(O(n^2)\) string longestPalindrome(string s) { int n = s.length(); string res; bool dp[n][n]; memset(dp, false, sizeof(dp)); int len = 0; for (int j = 0; j < n; j++) { for (int i = 0; i <= j; i++) { if (i == j) { // 長度為 1 dp[i][j] = true; } else if (j - i == 1) { // 長度為 2 dp[i][j] = s[i] == s[j]; } else { // 長度 > 2 dp[i][j] = s[i] == s[j] && dp[i+1][j-1]; } if (dp[i][j] && j - i + 1 > len) { // 比較長度 len = j - i + 1; res = s.substr(i, len); } } } return res; } 回目錄 Catalog ...

4. Median of Two Sorted Arrays Hardness: \(\color{red}\textsf{Hard}\) Ralated Topics: Array、Binary Search、Divide and Conquer 一、題目 Given two sorted arrays nums1 and nums2 of size m and n respectively, return the median of the sorted arrays. The overall run time complexity should be O(log (m+n)). Example 1: Input: nums1 = [1,3], nums2 = [2] Output: 2.00000 Explanation: merged array = [1,2,3] and median is 2. Example 2: Input: nums1 = [1,2], nums2 = [3,4] Output: 2.50000 Explanation: merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5. Constraints: ...

3. Longest Substring Without Repeating Characters Hardness: \(\color{orange}\textsf{Medium}\) Ralated Topics: Hash Table、String、Sliding Window 一、題目 Given a string s, find the length of the longest substring without repeating characters. Example 1: Input: s = “abcabcbb” Output: 3 Explanation: The answer is “abc”, with the length of 3. Example 2: Input: s = “bbbbb” Output: 1 Explanation: The answer is “b”, with the length of 1. Example 3: Input: s = “pwwkew” Output: 3 Explanation: The answer is “wke”, with the length of 3. Notice that the answer must be a substring, “pwke” is a subsequence and not a substring. Constraints: ...

2. Add Two Numbers Hardness: \(\color{orange}\textsf{Medium}\) Ralated Topics: Linked List、Math、Recursion 一、題目 You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order, and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list. You may assume the two numbers do not contain any leading zero, except the number 0 itself. Example 1: Input: l1 = [2,4,3], l2 = [5,6,4] Output: [7,0,8] Explanation: 342 + 465 = 807 Example 2: ...

難度分: 1748 這題的 minSize 和 maxSize 在 26 以內，範圍不會太大，可以用定長度的 sliding window 硬解。 class Solution { public: int maxFreq(string s, int maxLetters, int minSize, int maxSize) { int res = 0; for (int i = minSize; i <= maxSize; i++) { res = max(res, maxFreqWithWindowSize(s, maxLetters, i)); } return res; } int maxFreqWithWindowSize(string& s, int th, int k) { int n = s.size(); int cnt[26]; memset(cnt, 0, sizeof(cnt)); int uq = 0; int res = 0; unordered_map<string, int> map; for (int i = 0; i < k; i++) { if (cnt[s[i]-'a']++ == 0) uq++; } if (uq <= th) { map[s.substr(0, k)]++; res = 1; }; for (int i = k; i < n; i++) { if (cnt[s[i]-'a']++ == 0) uq++; if (--cnt[s[i-k]-'a'] == 0) uq--; if (uq <= th) { string t = s.substr(i-k+1, k); res = max(res, ++map[t]); } } return res; } };

這題要用 sliding window 解需要處理 circular iteration class Solution { public: vector<int> decrypt(vector<int>& code, int k) { int n = code.size(); vector<int> res(n, 0); if (k == 0) return res; int m = abs(k); int start = k > 0 ? 1 : n+k; int sum = 0; for (int i = 0; i < m; i++) { sum += code[(start + i) % n]; } res[0] = sum; for (int i = m, j = 1; i < m + n - 1; i++, j++) { int in = (start + i) % n; int out = (start + i - k) % n; sum += (code[(start + i) % n] - code[(start + i - m) % n]); res[j] = sum; } return res; } };