int sumRange(int i, int j) 返回数组 nums 中索引 left 和 right 之间的元素的 总和 ,包含 left 和 right 两点(也就是 nums[left] + nums[left + 1] + ... + nums[right] ) 示例1: 输入: ["NumArray", "sumRange", "sumRange", "sumRange"] [[[-2, 0, 3, -5, 2, -1]], [0,...
rangeSumQueryImmutable.printOriginArray(nums); rangeSumQueryImmutable.printPreSum(); rangeSumQueryImmutable.printRecoverArray(); System.out.println("\n"); System.out.println(rangeSumQueryImmutable.sumRange(2,2)); System.out.println(rangeSumQueryImmutable.sumRange(0,1)); System.out.println(rangeSu...
LeetCode 303. Range Sum Query - Immutable 程序员木子 香港浸会大学 数据分析与人工智能硕士在读Description 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...
int sumRange(int i, int j) 返回数组 nums 从索引 i 到 j(i ≤ j)范围内元素的总和,包含 i、j 两点(也就是 sum(nums[i], nums[i + 1], ... , nums[j])) 来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/range-sum-query-immutable 著作权归领扣网络所有。商业转载请联系官方...
sumRange(0, 2) -> 1 sumRange(2, 5) -> -1 sumRange(0, 5) -> -3 题意很简单,就是一个基本的求和问题,要是直接求的化,肯定超时,所以可以使用DP的思想。 代码如下: import java.util.ArrayList; import java.util.List; /* * 这里的求和部分使用的的是DP部分,不能直接求和,会超时的 ...
numArray.sumRange(0, 5); // return (-2) + 0 + 3 + (-5) + 2 + (-1) = -3 题目大意:求给定数组中,任意两个位置间所有数字的和 解题思路:求前缀和,然后用后面的减去前面的即可 classNumArray(object):def__init__(self,nums):""":type nums: List[int]"""sum_nums=[0]foriinrange...
LeetCode题目链接 题目: Given an integer arraynums, find the sum of the elements between indicesiandj(i≤j), inclusive. Example: Given nums = [-2, 0, 3, -5, 2, -1] sumRange(0, 2) -> 1 sumRange(2, 5) -> -1 sumRange(0, 5) -> -3 ...
思路一、Note 中指出,sumRange 会多次调用,所以构造函数O(n),sumRange函数O(1) classNumArray{public:vector<int>result;NumArray(vector<int>&nums){result.push_back(0);intn=nums.size();inttemp=0;for(inti=0;i<n;i++){temp=temp+nums[i];result.push_back(temp);}}intsumRange(inti,intj){...
计算索引 left 和 right (包含 left 和 right)之间的 nums 元素的 和 ,其中 left <= right实现 NumArray 类:
[LeetCode] Range Sum Query - Immutable 简介:The idea is fairly straightforward: create an array accu that stores the accumulated sum fornums such that accu[i] = nums[0] + . The idea is fairly straightforward: create an arrayaccuthat stores the accumulated sum fornumssuch thataccu[i] = ...