原题地址:https://oj.leetcode.com/problems/pascals-triangle-ii/ 题意: 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? 解题思路:思路同上一道题…

Pascal's Triangle Given numRows, generate the first numRows of Pascal's triangle. For example, given numRows = 5,Return [ [1], [1,1], [1,2,1], [1,3,3,1], [1,4,6,4,1] ] class Solution: # @return a list of lists of integers def generate(self, numRows):…

当我们需要改变数组的值时,如果从前往后遍历,有时会带来很多麻烦,比如需要插入值,导致数组平移,或者新的值覆盖了旧有的值,但旧有的值依然需要被使用.这种情况下,有时仅仅改变一下数组的遍历方向,就会避免这些困难. 最直观的一题是 剑指Offer上的面试题 4 另外一道例题,就是LeetCode上的 Pascal's Triangle II Pascal's Triangle II Given an index k, return the kth row of the Pascal's triangl…

Pascal's Triangle II Total Accepted: 19384 Total Submissions: 63446 My Submissions Question SolutionGiven 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 us…

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,…

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? 杨辉三角想必大家并不陌生,应该最早出现在初高中的数学中,其实就是二项式系数的一种写法. 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1…

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? 思路: 递归 package recursion; import java.util.List; import java.util.Arr…

Description: Given an index k, return the kth row of the Pascal's triangle. For example, given k = 3, Return [1,3,3,1]. public class Solution { public List<Integer> getRow(int rowIndex) { List<List<Integer>> list = new ArrayList<List&…

1.将tri初始化为[1],当rowIndex=0时,return的结果是:1,而题目要求应该是:[1],故将tri初始化为[[1]],返回结果设置为tri[0]即可满足要求: 2.最开始第二层循环是从1到i进行遍历,这样就不好控制数据的更新,因为更新第j个数据要用到原tri行的第j-1个数据,而此时第j-1个数据已经在上一轮被更新: 3.为了解决2中的问题,将第二层循环修改为从i到1进行遍历就可以了(因为第一个元素始终为1就不需要更新了) 4.函数用法:range(start,end,step…

1.这道题一次提交就AC了: 2.以前用C语言实现的话,初始化二维数组全部为0,然后每行第一个元素为1,只需要用a[i][j] = a[i-1][j]+a[i-1][j-1]就可以了: 3.在Python中难点应该就是每行的第一个元素和最后一个元素,最后一个元素通过判断j==i就可以区分了: class Solution: # @return a list of lists of integers def generate(self, numRows): ret = [] for i in ra…