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…

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<…

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. 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: You may assume that the array do…

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…

""" 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 """…

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…

#-*- 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]        "&…

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]…

The idea is fairly straightforward: create an array accu that stores the accumulated sum fornums such that accu[i] = nums[0] + ... + nums[i] in the initializer of NumArray. Then just return accu[j + 1] - accu[i] in sumRange. You may try the example i…

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…

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…

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  Note: You may assume that the array d…

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&…

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…

Question 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…

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 does not change. There are many calls to sumRange function. 求区域和,而且这个和也有可能被多次的调用,所以不能使用直接计算的方法,用dp解决,注意0这个点的特殊情况就…

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 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: 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: 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…

https://leetcode.com/problems/range-sum-query-immutable/ 用缓存撒 /** * @constructor * @param {number[]} nums */ var NumArray = function(nums) { this.nums = nums; this.cache = []; var acc = 0; for (var i = 0; i < nums.length; i++) { acc += nums[i]; this.…

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-] +…

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…