题目来源https://leetcode.com/problems/permutation-sequence/The set [1,2,3,…,n] contains a total of n! unique permutations.By listing and labeling all of the permutations in order,We get the following sequence (ie, for n = 3):"123""132""213""231""312""321" Given n and k, return the kth permutation sequence.题意分析Input: n, kOutput: the kth permutation in permutations based on nConditions: 返回第n...
Given an input string, reverse the string word by word.
For example,
Given s = "the sky is blue",
return "blue is sky the".Update (2015-02-12):
For C programmers: Try to solve it in-place in O(1) space.
linear time:
1. go through the string detect whether the place is ‘ ‘ if not append the char to word
2 if the place is ‘ ‘ and word is not "" append the word to res
3. go to the end append ...
# -*- coding: utf8 -*-‘‘‘__author__ = ‘dabay.wang@gmail.com‘https://oj.leetcode.com/problems/add-two-numbers/You are given two linked lists representing two non-negative numbers.The digits are stored in reverse order and each of their nodes contain a single digit.Add the two numbers and return it as a linked list.Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)Output: 7 -> 0 -> 8===Comments by Dabay===这...
op K Frequent ElementsGiven a non-empty array of integers, return the k most frequent elements.For example,Given [1,1,1,2,2,3] and k = 2, return [1,2].Note: You may assume k is always valid, 1 ≤ k ≤ number of unique elements.Your algorithm‘s time complexity must be better than O(n log n), where n is the array‘s size. 求出现频率最高的K个数，要求时间复杂度不能超过 O(n log n)。选择最简单的快排...
# 解题思路: # 创建一个字典映射关系 dicts# 使用一个栈stk 遍历字符串s 得到一个新的字符串curItem 如果lastItem在dicts中的value和它相等 不做任何操作# 如果不等 入栈 有lastItem的 先append lastItem 然后是curItem ## 最后判断如果stk为空说明所给字符串匹配 return true class Solution(object):def isValid(self, s):""":type s: str:rtype: bool"""strLen = len(s)if strLen <= 1:return Falsedicts={"(":")","{":"}","[":...
题目来源： https://leetcode.com/problems/search-insert-position/ 题意分析： 给定一个排好序的数组和一个target，如果target在数组里面，那么返回他的位置，否者返回他应该插入哪个位置。 题目思路： 这也是一个标准的二分查找。如果没有找到，那么和first和last位置的数比较一下就可以得到答案。 代码（python）： 1class Solution(object):2def searchInsert(self, nums, target):3""" 4 :type nums: List[i...
题目https://leetcode-cn.com/problems/minimum-number-of-frogs-croaking/给你一个字符串 croakOfFrogs，它表示不同青蛙发出的蛙鸣声（字符串 "croak" ）的组合。由于同一时间可以有多只青蛙呱呱作响，所以?croakOfFrogs 中会混合多个 “croak” 。请你返回模拟字符串中所有蛙鸣所需不同青蛙的最少数目。注意：要想发出蛙鸣 "croak"，青蛙必须 依序 输出 ‘c’, ’r’, ’o’, ’a’, ’k’ 这 5 个字母。如果没有输出全部五个字母...
Given an n-ary tree, return the postorder traversal of its nodes‘ values. For example, given a 3-ary tree: Return its postorder traversal as: [5,6,3,2,4,1]. Note: Recursive solution is trivial, could you do it iteratively? Recursion Solution:"""
# Definition for a Node.
class Node(object):def __init__(self, val, children):self.val = valself.children = children
"""
class Solution(object):def posto...
原题地址：https://oj.leetcode.com/problems/first-missing-positive/题意：Given an unsorted integer array, find the first missing positive integer.For
example,Given [1,2,0] return 3,and [3,4,-1,1] return 2.Your algorithm should run in O(n) time and uses
constant space.解题思路：要求时间复杂度为O(n)和常数空间。又是一个tricky的解法。class Solution:# @param A, a list of integers# @return an integer# ...
Given a list of words (without duplicates), please write a program that returns all concatenated words in the given list of words.A concatenated word is defined as a string that is comprised entirely of at least two shorter words in the given array.Example:Input: ["cat","cats","catsdogcats","dog","dogcatsdog","hippopotamuses","rat","ratcatdogcat"]Output: ["catsdogcats","dogcatsdog","ratcatdogcat"]...
思路是这样的，我们从第一个字符开始向后依次找，直到找到一个断句的地方，使得当前获得的子串在dict中，若找到最后都没找到，那么就是False了。在找到第一个后，接下来找下一个断句处，当然是从第一个断句处的下一个字符开始找连续的子串，但是这时与第一个就稍有不同，比如说word=‘ab’， dict={ ‘a‘, ab‘, ...},在找到a后，接下来处理的是b，我们发现b不在dict中，但是我们发现b可以和a结合，形成ab，而ab在dict中，所以这里...
题目来源： https://leetcode.com/problems/search-a-2d-matrix/ 题意分析： 给一个m×n矩阵，矩阵是按照从小到大排列，也就是a[i][j]和a[m][n]如果i > m则a[i][j]>a[m][n]，如果i == m，j > n,则a[i][j]>a[m][n]。 题目思路： 利用二分查找，首先确定在那个行，然后继续二分查找。 代码（Python）： 1class Solution(object):2def searchMatrix(self, matrix, target):3""" 4 :type matrix: List[List[int]]5 ...
原题地址：https://oj.leetcode.com/problems/powx-n/题意：Implement
pow(x, n).解题思路：求幂函数的实现。使用递归，类似于二分的思路，解法来自Mark Allen Weiss的《数据结构与算法分析》。代码：class Solution:# @param x, a float# @param n, a integer# @return a floatdef pow(self, x, n):if n == 0:return 1elif n == 1:return xelif n % 2:return pow(x*x,n/2)*xelse:return pow(x*x,n/2) 原文：http://www.cnblogs.c...
【python】Leetcode每日一题-寻找旋转排序数组中的最小元素2【题目描述】已知一个长度为 n 的数组，预先按照升序排列，经由 1 到 n 次 旋转 后，得到输入数组。例如，原数组nums = [0,1,2,4,5,6,7]在变化后可能得到：
若旋转4次，则可以得到 [4,5,6,7,0,1,2]
若旋转4次，则可以得到 [0,1,2,4,5,6,7]
注意，数组[a[0], a[1], a[2], ..., a[n-1]]旋转一次 的结果为数组[a[n-1], a[0], a[1], a[2], ..., a[n-2]]。给你一个可能存在重复...
简单的插入排序，总是超时，暂且放在这记录一下。class Solution:# @param head, a ListNode# @return a ListNodedef insertionSortList(self, head):if head == None or head.next == None:return headpsuhead = ListNode(-1)while head:tail = psuheadheadnext = head.nextwhile tail.next and tail.next.val < head.val:tail = tail.nexthead.next = tail.nexttail.next = headhead = headnextreturn psuhead.next
Last execute...