#-*- coding: UTF-8 -*-#杨辉三角返回给定行#方法:自上而下考虑问题,从给定的一行计算出下一行,则给定行数之后,计算的最后一行就是求解的最后一行class Solution(object):    def getRow(self, rowIndex):        """        :type rowIndex: int        :rtype: List[int]        """        #初始化杨辉三…

#-*- coding: UTF-8 -*-#杨辉三角class Solution(object):    def generate(self, numRows):        """        :type numRows: int        :rtype: List[List[int]]        """        result=[];row=1        while True:            tmplist=[1…

一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 Given an index k, return the kth row of the Pascal's triangle. For example, given k = 3, Return [1,3,3,1]. Note: Could you optimize your algorithm to use onl…

题目: Given an index k, return the kth row of the Pascal's triangle. For example, given k = 3,Return [1,3,3,1]. Note:Could you optimize your algorithm to use only O(k) extra space? 提示: 此题要求给出杨辉三角对应行号下的所有数字序列,且空间复杂度为O(n).这里我们可以根据杨辉三角的定义,逐行计算.顺序从左到右或者从右到…

Description Given a non-negative index k where k ≤ 33, return the kth index row of the Pascal's triangle. Note that the row index starts from 0. In Pascal's triangle, each number is the sum of the two numbers directly above it. Example: Input: Output…

Given an index k, return the kth row of the Pascal's triangle. For example, given k = 3,Return [1,3,3,1]. Note:Could you optimize your algorithm to use only O(k) extra space? [思路] 我们为了满足空间复杂度的要求,我们新建两个ArrayList,一个负责存储上一个Pascal行的结果,一个根据上一个Pascal行得出当前P…

118. Pascal's Triangle 第一种解法:比较麻烦 https://leetcode.com/problems/pascals-triangle/discuss/166279/cpp-beats-1002018.9.3(with-annotation) class Solution { public: vector<vector<int>> generate(int numRows) { vector<vector<int>> result;…

原文题目: 118. Pascal's Triangle 119. Pascal's Triangle II 读题: 杨辉三角问题 '''118''' class Solution(object): def generate(self, numRows): """ :type numRows: int :rtype: List[List[int]] """ if not numRows: return [] result = [] for i i…

Given a non-negative index k where k ≤ 33, return the kth index row of the Pascal's triangle. Note that the row index starts from 0. Example: Input: 3 Output: [1,3,3,1]  原题地址:Pascal's Triangle II 题意: 杨辉三角 代码:  class Solution(object): def getRow(self,…

题目来源 https://leetcode.com/problems/pascals-triangle-ii/ Given an index k, return the kth row of the Pascal's triangle. For example, given k = 3,Return [1,3,3,1]. 题意分析 Input:integer Output:kth row of the Pascal's triangle Conditions:只返回第n行 题目思路 同118,不…

Given a non-negative index k where k ≤ 33, return the kth index row of the Pascal's triangle. Note that the row index starts from 0. In Pascal's triangle, each number is the sum of the two numbers directly above it. Example: Input: 3 Output: [1,3,3,1…

Given an index k, return the kth row of the Pascal's triangle. For example, given k = 3,Return [1,3,3,1]. Note:Could you optimize your algorithm to use only O(k) extra space? 118. Pascal's Triangle 的拓展,给一个索引k,返回杨辉三角的第k行. 解法:题目要求优化到 O(k) 的空间复杂,那么就不能把每…

#-*- coding: UTF-8 -*- #既然不能使用加法和减法,那么就用位操作.下面以计算5+4的例子说明如何用位操作实现加法:#1. 用二进制表示两个加数,a=5=0101,b=4=0100:#2. 用and(&)操作得到所有位上的进位carry=0100;#3. 用xor(^)操作找到a和b不同的位,赋值给a,a=0001:#4. 将进位carry左移一位,赋值给b,b=1000:#5. 循环直到进位carry为0,此时得到a=1001,即最后的sum.#!!!!!!关于负数的运算.…

Given an index k, return the kth row of the Pascal's triangle. For example, given k = 3,Return [1,3,3,1]. Note:Could you optimize your algorithm to use only O(k) extra space? 题目标签:Array 这道题目与之前那题不同的地方在于,之前是给我们一个行数n,让我们把这几行的全部写出来,这样就可以在每写新的一行的时候根据之前的那…

Problem: Given an index k, return the kth row of the Pascal's triangle. For example, given k = 3,Return [1,3,3,1]. Note:Could you optimize your algorithm to use only O(k) extra space? Summary: 返回杨辉三角(帕斯卡三角)的第k行. Solution: 1. 若以二维数组的形式表示杨辉三角,则可轻易推算出ro…

Given an index k, return the kth row of the Pascal's triangle. For example, given k = 3,Return [1,3,3,1]. Note:Could you optimize your algorithm to use only O(k) extra space? 上一道题的延伸版,就是直接求出第k行的数,要求用o(k)的空间复杂度. 也是直接相加就可以了. public class Solution { pub…

题目描述: Given an index k, return the kth row of the Pascal's triangle. For example, given k = 3,Return [1,3,3,1]. 解题思路: 每次在上一个list前面插入1,然后后面的每两个间相加赋值给前一个数. 代码描述: public class Solution { public List<Integer> getRow(int rowIndex) { List<Integer> r…

题目是: Given an index k, return the kth row of the Pascal's triangle. For example, given k = 3,Return [1,3,3,1]. Note:Could you optimize your algorithm to use only O(k) extra space? (注意:这里要求空间为O(k)) 一个满足条件的答案如下: public class Solution { public IList<int…

Given a non-negative integer numRows, generate the first numRows of Pascal's triangle. In Pascal's triangle, each number is the sum of the two numbers directly above it. Example: Input: 5 Output: [ [1], [1,1], [1,2,1], [1,3,3,1], [1,4,6,4,1] ] class…

Given a non-negative index k where k ≤ 33, return the kth index row of the Pascal's triangle. Note that the row index starts from 0. In Pascal's triangle, each number is the sum of the two numbers directly above it. Example: Input: 3 Output: [1,3,3,1…

Given an index k, return the kth row of the Pascal's triangle. For example, given k = 3, Return [1,3,3,1]. 解题思路: 注意,本题的k相当于上题的k+1,其他照搬即可,JAVA实现如下: public List<Integer> getRow(int rowIndex) { List<Integer> alist=new ArrayList<Integer>();…

1. 题目 1.1 英文题目 Given an integer rowIndex, return the rowIndexth (0-indexed) row of the Pascal's triangle. In Pascal's triangle, each number is the sum of the two numbers directly above it as shown: 1.2 中文题目 输出杨辉三角形的指定行 1.3输入输出 输入 输出 rowIndex = 3 [1,3…

Given an index k, return the kth row of the Pascal's triangle. For example, given k = 3,Return [1,3,3,1]. 分析 二项式展开式的系数 不加(long long)提交会出错. class Solution { public: vector<int> getRow(int rowIndex) { vector<int> vals; vals.resize(rowIndex+); va…

Given an index k, return the kth row of the Pascal's triangle. For example, given k = 3, Return [1,3,3,1]. Note: Could you optimize your algorithm to use only O(k) extra space?   思路:从右往左遍历并填写三角形的图结构 class Solution { public: vector<int> getRow(int ro…

#-*- coding: UTF-8 -*- #AC源码[意外惊喜,还以为会超时]class Solution(object):    def twoSum(self, nums, target):        """        :type nums: List[int]        :type target: int        :rtype: List[int]        """         for i in xrange(…

#-*- coding: UTF-8 -*-#利用strip函数去掉字符串去除空格(其实是去除两边[左边和右边]空格)#利用split分离字符串成列表class Solution(object):    def lengthOfLastWord(self, s):        """        :type s: str        :rtype: int        """        if s==None:return 0     …

#-*- coding: UTF-8 -*-#需要考虑多种情况#以下几种是可以返回的数值#1.以0开头的字符串,如01201215#2.以正负号开头的字符串,如'+121215':'-1215489'#3.1和2和空格混合形式[顺序只能是正负号-0,空格位置可以随意]的:'+00121515'#4.正数小于2147483647,负数大于-2147483648的数字#其他的情况都是返回0,因此在判断 是把上述可能出现的情况列出来,其他的返回0#AC源码如下class Solution(object…

#-*- coding: UTF-8 -*-class Solution(object):    def compareVersion(self, version1, version2):        """        :type version1: str        :type version2: str        :rtype: int        """        versionl1=version1.split('.'…

class Solution(object):    def convertToTitle(self, n):        """        :type n: int        :rtype: str        """        res=''        while n>0:            tmp=n            n=(n-1)/26            res+=chr(65+(tmp-1)%26)…

#-*- coding: UTF-8 -*-#2147483648#在32位操作系统中,由于是二进制,#其能最大存储的数据是1111111111111111111111111111111.#正因为此,体现在windows或其他可视系统中的十进制应该为2147483647.#32位数的范围是 -2147483648~2147483648class Solution(object):    def reverse(self, x):        """        :type…