Given an array of integers nums, write a method that returns the "pivot" index of this array. We define the pivot index as the index where the sum of the numbers to the left of the index is equal to the sum of the numbers to the right of the ind…

leetcode 724. Find Pivot Index 题目描述:在数组中找到一个值,使得该值两边所有值的和相等.如果值存在,返回该值的索引,否则返回-1 思路:遍历两遍数组,第一遍求出数组的和,第二遍开始,保存左边所有的值的和,当左边值的和的2倍加上当前值等于数组和时,就是要找的索引.时间复杂度为o(n),空间复杂度为o(1). class Solution(object): def pivotIndex(self, nums): """ :type nums: Li…

problem 724. Find Pivot Index 题意:先求出数组的总和,然后维护一个当前数组之和curSum,然后对于遍历到的位置,用总和减去当前数字,看得到的结果是否是curSum的两倍,是的话,那么当前位置就是中枢点,返回即可:否则就将当前数字加到curSum中继续遍历,遍历结束后还没返回,说明没有中枢点,返回-1即可. solution: 注意,不管有几种方式,最后的结果都是从最左边开始的那个结果,那么我们就直接从最左边开始计算,返回第一个满足条件的即可. class Solu…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 先求和,再遍历 日期 题目地址:https://leetcode.com/problems/find-pivot-index/description/ 题目描述 Given an array of integers nums, write a method that returns the "pivot" index of this arr…

Given an array of integers nums, write a method that returns the "pivot" index of this array. We define the pivot index as the index where the sum of the numbers to the left of the index is equal to the sum of the numbers to the right of the ind…

Given an array of integers nums, write a method that returns the "pivot" index of this array. We define the pivot index as the index where the sum of the numbers to the left of the index is equal to the sum of the numbers to the right of the ind…

［抄题］: Given an array of integers nums, write a method that returns the "pivot" index of this array. We define the pivot index as the index where the sum of the numbers to the left of the index is equal to the sum of the numbers to the right of t…

最近,在用解决LeetCode问题的时候,做了349: Intersection of Two Arrays这个问题,就是求两个列表的交集.我这种弱鸡,第一种想法是把问题解决,而不是分析复杂度,于是写出了如下代码: class Solution(object): def intersection(self, nums1, nums2): """ :type nums1: List[int] :type nums2: List[int] :rtype: List[int] &q…

题目描述:把一个二维数组顺时针旋转90度: 思路: 对于数组每一圈进行旋转,使用m控制圈数: 每一圈的四个元素顺时针替换,可以直接使用Python的解包,使用k控制每一圈的具体元素: class Solution(object): def rotate(self, matrix): """ :type matrix: List[List[int]] :rtype: void Do not return anything, modify matrix in-place inst…

用一个整型数组构建一个二叉树,根结点是数组中的最大值,左右子树分别是根结点的值在数组中左右两边的部分. 分析,这是二叉树中比较容易想到的问题了,直接使用递归就行了,代码如下: class Solution(object): def constructMaximumBinaryTree(self, nums): """ :type nums: List[int] :rtype: TreeNode """ if not nums: return No…