标签:pytho UNC substr ges str ret turn tools mamicode
给定一个包含 n 个整数的数组 nums,判断 nums 中是否存在三个元素 a,b,c ,使得 a + b + c = 0 ?找出所有满足条件且不重复的三元组。
注意:答案中不可以包含重复的三元组。
例如, 给定数组 nums = [-1, 0, 1, 2, -1, -4],
满足要求的三元组集合为:
[
[-1, 0, 1],
[-1, -1, 2]
]
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/3sum
class Solution(object): def threeSum(self, nums): """ :type nums: List[int] :rtype: List[List[int]] """ def double_pointer_search(left, right, first_num, nums, result): while left < right: if 0 < left < right and nums[left] == nums[left - 1] or first_num + nums[left] + nums[right] < 0: left += 1 if left < right < len(nums)-1 and nums[right] == nums[right + 1] or first_num + nums[left] + nums[right] > 0: right -= 1 else: result.append([first_num, nums[left], nums[right]]) left += 1 right -= 1 result = [] if len(nums) < 3: return result nums.sort() n = len(nums) for i in range(n - 2): if nums[i] + nums[i + 1] + nums[i + 2] > 0: break if nums[i] + nums[-1] + nums[-2] < 0: continue if i > 0 and nums[i] == nums[i - 1]: continue double_pointer_search(i + 1, n - 1, nums[i], nums, result) return result
给定一组非负整数,重新排列它们的顺序使之组成一个最大的整数。
示例 1:
输入: [10,2]
输出: 210
示例 2:
输入: [3,30,34,5,9]
输出: 9534330
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/largest-number
from functools import cmp_to_key class Solution(object): def largestNumber(self, nums): """ :type nums: List[int] :rtype: str """ def new_sort(num1, num2): if num1 + num2 > num2 + num1: return -1 elif num1 + num2 < num2 + num1: return 1 return 0 result = "".join(sorted(list(map(lambda num: str(num), nums)), key=cmp_to_key(lambda x, y: new_sort(x, y)))) while result.startswith("0") and len(result) > 1: result = result[1:] return result
给定两个由一些闭区间组成的列表,每个区间列表都是成对不相交的,并且已经排序。
返回这两个区间列表的交集。
(形式上,闭区间 [a, b](其中 a <= b)表示实数 x 的集合,而 a <= x <= b。两个闭区间的交集是一组实数,要么为空集,要么为闭区间。例如,[1, 3] 和 [2, 4] 的交集为 [2, 3]。)
示例:
输入:A = [[0,2],[5,10],[13,23],[24,25]], B = [[1,5],[8,12],[15,24],[25,26]]
输出:[[1,2],[5,5],[8,10],[15,23],[24,24],[25,25]]
注意:输入和所需的输出都是区间对象组成的列表,而不是数组或列表。
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/interval-list-intersections
class Solution(object): def intervalIntersection(self, A, B): """ :type A: List[List[int]] :type B: List[List[int]] :rtype: List[List[int]] """ result = [] i, j = 0, 0 while i < len(A) and j < len(B): left, right = max(A[i][0], B[j][0]), min(A[i][1], B[j][1]) if left <= right: result.append([left, right]) if A[i][1] < B[j][1]: i += 1 else: j += 1 return result
给定一个由 0 和 1 组成的矩阵,找出每个元素到最近的 0 的距离。
两个相邻元素间的距离为 1 。
示例 1:
输入:
0 0 0
0 1 0
0 0 0
输出:
0 0 0
0 1 0
0 0 0
示例 2:
输入:
0 0 0
0 1 0
1 1 1
输出:
0 0 0
0 1 0
1 2 1
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/01-matrix
class Solution(object): def updateMatrix(self, matrix): """ :type matrix: List[List[int]] :rtype: List[List[int]] """ def bfs(i, j): que, distance, visited = [(i, j)], 0, set() while que: distance += 1 new_que = [] for new_point in que: for i, j in [(0, -1), (0, 1), (-1, 0), (1, 0)]: new_i, new_j = new_point[0] + i, new_point[1] + j if 0 <= new_i < len(matrix) and 0 <= new_j < len(matrix[0]) and (new_i,new_j) not in visited: if matrix[new_i][new_j] != 0: new_que.append((new_i, new_j)) visited.add((new_i, new_j)) else: return distance que = new_que return distance new_matrix = [[0 for i in range(len(matrix[0]))] for j in range(len(matrix))] for row_index, row in enumerate(matrix): for col_index, point in enumerate(row): if matrix[row_index][col_index] != 0: new_matrix[row_index][col_index] = bfs(row_index, col_index) return new_matrix
class Solution(object): def searchRange(self, nums, target): """ :type nums: List[int] :type target: int :rtype: List[int] """ if not nums: return [-1, -1] left, right = 0, len(nums) - 1 while left + 1 < right: mid = left + (right - left) / 2 if nums[mid] == target: i = mid - 1 while i >= 0 and nums[i] == target: i -= 1 j = mid + 1 while j < len(nums) and nums[j] == target: j += 1 return [i + 1, j - 1] if nums[mid] > target: right = mid elif nums[mid] < target: left = mid if nums[left] == target and nums[right] == target: return [left, right] if nums[right] == target: return [right, right] if nums[left] == target: return [left, left] return [-1, -1]
给定一个非空字符串 s 和一个包含非空单词列表的字典 wordDict,在字符串中增加空格来构建一个句子,使得句子中所有的单词都在词典中。返回所有这些可能的句子。
说明:
分隔时可以重复使用字典中的单词。
你可以假设字典中没有重复的单词。
示例 1:
输入:
s = "catsanddog"
wordDict = ["cat", "cats", "and", "sand", "dog"]
输出:
[
"cats and dog",
"cat sand dog"
]
示例 2:
输入:
s = "pineapplepenapple"
wordDict = ["apple", "pen", "applepen", "pine", "pineapple"]
输出:
[
"pine apple pen apple",
"pineapple pen apple",
"pine applepen apple"
]
解释: 注意你可以重复使用字典中的单词。
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/word-break-ii
class Solution(object): def wordBreak(self, s, wordDict): """ :type s: str :type wordDict: List[str] :rtype: List[str] """ def dfs(s, word_dict, sets): if s in sets: return sets[s] if len(s) == 0: return [] partitions = [] if s in word_dict: partitions.append(s) for i in range(1, len(s)): word = s[:i] if word not in word_dict: continue sub_partitions = dfs(s[i:], word_dict, sets) partitions.extend(map(lambda sub_partition: word + " " + sub_partition, sub_partitions)) # for sub_partition in sub_partitions: # partitions.append(word + " " + sub_partition) sets[s] = partitions return partitions return dfs(s, set(wordDict), {})
给定一个非空字符串 s 和一个包含非空单词列表的字典 wordDict,在字符串中增加空格来构建一个句子,使得句子中所有的单词都在词典中。返回所有这些可能的句子。
说明:
分隔时可以重复使用字典中的单词。
你可以假设字典中没有重复的单词。
示例 1:
输入:
s = "catsanddog"
wordDict = ["cat", "cats", "and", "sand", "dog"]
输出:
[
"cats and dog",
"cat sand dog"
]
示例 2:
输入:
s = "pineapplepenapple"
wordDict = ["apple", "pen", "applepen", "pine", "pineapple"]
输出:
[
"pine apple pen apple",
"pineapple pen apple",
"pine applepen apple"
]
解释: 注意你可以重复使用字典中的单词。
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/word-break-ii
class Solution(object): def longestPalindrome(self, s): """ :type s: str :rtype: str """ if not s: return "" n, start, longest = len(s), 0, 1 f = [[False]*n for _ in range(n)] for i in range(n-1, -1, -1): f[i][i] = True for j in range(i + 1, n): f[i][j] = s[i] == s[j] and (j - i < 2 or f[i+1][j-1]) if f[i][j] and longest < j - i + 1: longest = j - i + 1 start = i return s[start:start + longest]
给定一个链表,旋转链表,将链表每个节点向右移动 k 个位置,其中 k 是非负数。
示例 1:
输入: 1->2->3->4->5->NULL, k = 2
输出: 4->5->1->2->3->NULL
解释:
向右旋转 1 步: 5->1->2->3->4->NULL
向右旋转 2 步: 4->5->1->2->3->NULL
示例 2:
输入: 0->1->2->NULL, k = 4
输出: 2->0->1->NULL
解释:
向右旋转 1 步: 2->0->1->NULL
向右旋转 2 步: 1->2->0->NULL
向右旋转 3 步: 0->1->2->NULL
向右旋转 4 步: 2->0->1->NULL
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/rotate-list
class Solution(object): def rotateRight(self, head, k): """ :type head: ListNode :type k: int :rtype: ListNode """ if not head: return head cur_node = head size = self.calculate_size(cur_node) k = k % size if k == 0: return head cur_node = head new_head = self.get_new_head(cur_node, k, size) cur_node = new_head self.attach_two_linked_list(cur_node, head) return new_head def attach_two_linked_list(self, cur_node, head): while cur_node.next: cur_node = cur_node.next cur_node.next = head def get_new_head(self, cur_node, k, size): len = 1 while len < size - k: len += 1 cur_node = cur_node.next new_head = cur_node.next cur_node.next = None return new_head def calculate_size(self, node): size = 0 while node != None: size += 1 node = node.next return size
标签:pytho UNC substr ges str ret turn tools mamicode
原文地址:https://www.cnblogs.com/rnanprince/p/12180460.html