[Python] Chaining Comparison Operators (a > b and b > c → a > b > c)
- 2023.08.10
- Python
a > b and b > c→a > b > c
a > b and b > c→a > b > c
https://leetcode.com/problems/search-in-rotated-sorted-array-ii Solution Leetcode # 33. Search in Rotated Sorted Array 的 solution 配合題目修改了 input 跟 output class Solution: def search(self, nums: List[int ...
https://leetcode.com/problems/minimize-the-maximum-difference-of-pairs Solution: Greedy Method + Binary Search Greedy Method – count_valid_pairs(threshold) count_valid_pairs(threshold): 事先已將num ...
https://leetcode.com/problems/search-in-rotated-sorted-array Solution n := len(nums) nums = [numk, numk+1, …numn-1] + [num0, num1, …, numk-1] = numsA + numsB numi < numj, if i < j; 0 ...
https://leetcode.com/problems/encode-and-decode-strings Solution Time Complexity: O(len(s)) Space Complexity: O(len(strs)) (The input and output generally do not count towards the space complexity.) c ...
https://leetcode.com/problems/longest-common-subsequence Solution: Top-Down DP Time Complexity: O(len(text1) * len(text2)) Space Complexity: O(len(text1) * len(text2)) (The input and output generally ...
https://leetcode.com/problems/maximum-score-from-performing-multiplication-operations Solution: Top-Down DP m := len(multipliers) n := len(nums) (end – start) = n – 1, n -2, …, n ...
https://leetcode.com/problems/house-robber-ii Solution _rob() := 必定搶最後一間的收益 結果 := max(搶最後一間, 不搶最後一間) = max(搶最後一間 and 不搶第一間, 不搶最後一間) = max(_rob(nums[1:]), robs(nums[:-1])) Time Complexity: O ...
https://leetcode.com/problems/delete-and-earn Solution Time Complexity: O(len(nums)) Space Complexity: O(len(counter)) (The input and output generally do not count towards the space complexity.) class ...
https://leetcode.com/problems/n-th-tribonacci-number Solution Time Complexity: O(n) Space Complexity: O(n) (The input and output generally do not count towards the space complexity.) class Solution: d ...