303. Range Sum Query - Immutable class NumArray { private: vector<int> v; public: NumArray(vector<int> &nums) { int n = nums.size(); v = vector<int>(n + 1); int sum = 0; for (int i = 1; i <= n; ++i) { sum += nums[i - 1]; v[i] = su…
Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive. Example: Given nums = [-2, 0, 3, -5, 2, -1] sumRange(0, 2) -> 1 sumRange(2, 5) -> -1 sumRange(0, 5) -> -3 Note: You may assume that the array do…
Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive. Example: Given nums = [-2, 0, 3, -5, 2, -1] sumRange(0, 2) -> 1 sumRange(2, 5) -> -1 sumRange(0, 5) -> -3 Note: 1: You may assume that the array…
题意:查询一个数组在(i,j]范围内的元素的和. 思路非常简单,做个预处理,打个表就好 拓展:可以使用树状数组来完成该统计,算法复杂度为(logn),该数据结构强力的地方是实现简单,而且能完成实时更新以及上面的统计和 class NumArray { public: vector<int> sum; NumArray(vector<int> &nums) { sum.push_back(); ; i< nums.size(); ++i){ int m = sum[i]…
题目描述: Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive. Example: Given nums = [-2, 0, 3, -5, 2, -1] sumRange(0, 2) -> 1 sumRange(2, 5) -> -1 sumRange(0, 5) -> -3 Note: You may assume that the ar…
题目: Given an integer array nums, find the sum of the elements between indices iand j (i ≤ j), inclusive. Example: Given nums = [-2, 0, 3, -5, 2, -1] sumRange(0, 2) -> 1 sumRange(2, 5) -> -1 sumRange(0, 5) -> -3 Note: You may assume that the array…
翻译 给定一个整型数组nums,找出在索引i到j(i小于等于j)之间(包含i和j)的全部元素之和. 比如: 给定nums = [-2,0,3,-5,2,-1] sumRange(0, 2) -> 1 sumRange(2, 5) -> -1 sumRange(0, 5) -> -3 批注: 你能够假定这个数组不会改变. 这里会有非常多次对sumRange函数的调用. 原文 Given an integer array nums, find the sum of the elements…
Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive. Example: Given nums = [-2, 0, 3, -5, 2, -1] sumRange(0, 2) -> 1 sumRange(2, 5) -> -1 sumRange(0, 5) -> -3 Note: You may assume that the array do…
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 解题方法 保存累积和 日期 题目地址:https://leetcode.com/problems/range-sum-query-immutable/description/ 题目描述 Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), in…
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 我的个人博客已创建,欢迎大家持续关注! 一天一道leetcode系列依旧在csdn上继续更新,除此系列以外的文章均迁移至我的个人博客 另外,本系列文章已整理并上传至gitbook,网址:点我进 欢迎转载,转载请注明出处! (一)题目 Given an integer array nums, find the sum of the elements b…
#-*- coding: UTF-8 -*- #Tags:dynamic programming,sumRange(i,j)=sum(j)-sum(i-1)class NumArray(object):    sums=[]    def __init__(self, nums):        """        initialize your data structure here.        :type nums: List[int]        "&…
问题: Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive. Example: Given nums = [-2, 0, 3, -5, 2, -1] sumRange(0, 2) -> 1 sumRange(2, 5) -> -1 sumRange(0, 5) -> -3 Note: You may assume that the arra…
Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive. Example: Given nums = [-2, 0, 3, -5, 2, -1] sumRange(0, 2) -> 1 sumRange(2, 5) -> -1 sumRange(0, 5) -> -3 Note: 1: You may assume that the array…
Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive. Example: Given nums = [-2, 0, 3, -5, 2, -1] sumRange(0, 2) -> 1 sumRange(2, 5) -> -1 sumRange(0, 5) -> -3 Note: You may assume that the array do…
题目: Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive. Example: Given nums = [-2, 0, 3, -5, 2, -1] sumRange(0, 2) -> 1 sumRange(2, 5) -> -1 sumRange(0, 5) -> -3 Note: You may assume that the arra…
Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive. Example: Given nums = [-2, 0, 3, -5, 2, -1] sumRange(0, 2) -> 1 sumRange(2, 5) -> -1 sumRange(0, 5) -> -3 法一:暴力 class NumArray { public: vector&…
给定一个数组,求出数组从索引 i 到 j  (i ≤ j) 范围内元素的总和,包含 i,  j 两点.例如:给定nums = [-2, 0, 3, -5, 2, -1],求和函数为sumRange()sumRange(0, 2) -> 1sumRange(2, 5) -> -1sumRange(0, 5) -> -3注意:    你可以假设数组不可变.    会多次调用 sumRange 方法.详见:https://leetcode.com/problems/range-sum-quer…
Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive. Example: Given nums = [-2, 0, 3, -5, 2, -1] sumRange(0, 2) -> 1 sumRange(2, 5) -> -1 sumRange(0, 5) -> -3 Note: You may assume that the array do…
1.这道题目与pat中的1046. Shortest Distance (20)相类似: 2.使用一个数组dp[i],记录0到第i个数的和 3.求i到j之间的和时,输出dp[j]-dp[i]+num[i]即可. AC代码如下: class NumArray { public: vector<int> dp; vector<int> num; NumArray(vector<int> &nums) { int n=nums.size(); dp=vector<…
Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2). The above rectangle (with the red border) is defined by (row1, col1) = (2, 1) and (row2, co…
303. Range Sum Query - Immutable Easy Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive. Example: Given nums = [-2, 0, 3, -5, 2, -1] sumRange(0, 2) -> 1 sumRange(2, 5) -> -1 sumRange(0, 5) -> -3…
Range Sum Query - Immutable Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive. Example: Given nums = [-2, 0, 3, -5, 2, -1] sumRange(0, 2) -> 1 sumRange(2, 5) -> -1 sumRange(0, 5) -> -3 Note: You…
Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive. The update(i, val) function modifies nums by updating the element at index i to val. Example: Given nums = [1, 3, 5] sumRange(0, 2) -> 9 update(1, 2…
Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive. Example: Given nums = [-2, 0, 3, -5, 2, -1] sumRange(0, 2) -> 1 sumRange(2, 5) -> -1 sumRange(0, 5) -> -3 Note: You may assume that the array do…
Immutable [抄题]: Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive. Example: Given nums = [-2, 0, 3, -5, 2, -1] sumRange(0, 2) -> 1 sumRange(2, 5) -> -1 sumRange(0, 5) -> -3 [暴力解法]: 时间分析:n 空间分析:n…
Range Sum Query - Mutable 题解 原创文章,拒绝转载 题目来源:https://leetcode.com/problems/range-sum-query-mutable/description/ Description Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive. The update(i, val) functi…
原题链接在这里:https://leetcode.com/problems/range-sum-query-2d-mutable/ 题目: Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2). The above rectangle (…
https://leetcode.com/problems/range-sum-query-immutable/ class NumArray { public: vector<int> vec; public: NumArray(vector<int> &nums) { if(nums.empty()) return; else { vec.push_back(nums[]); ;i<nums.size();++i) vec.push_back(vec[i-] +…
题目 Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive. Example: Given nums = [-2, 0, 3, -5, 2, -1] sumRange(0, 2) -> 1 sumRange(2, 5) -> -1 sumRange(0, 5) -> -3 Note: You may assume that the array…
这是悦乐书的第204次更新,第214篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第70题(顺位题号是303).给定整数数组nums,找到索引i和j(i≤j)之间的元素之和,包括端点.例如: 给定nums = [-2,0,3,-5,2,-1] sumRange(0,2) - > 1 sumRange(2,5) - > -1 sumRange(0,5) - > -3 注意: 您可以假设数组不会更改. sumRange函数有很多调用. 本次解题使用的开发工具是e…