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…

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. 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. 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. 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…

Problem: 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 up…

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…

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 (…

Range Sum Query - Mutable 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] sumRa…

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 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…

一. 题目描写叙述 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 u…

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…

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…

题目: 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(…

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…

题目 题目 思路 一看就是单点更新和区间求和,故用线段树做. 一开始没搞清楚,题目给定的i是从0开始还是从1开始,还以为是从1开始,导致后面把下标都改掉了,还有用区间更新的代码去实现单点更新,虽然两者思路是一样的,但是导致TLE,因为区间会把所有都递归一遍,加了个判断,就ok了. if (idx <= middle) { this->updateHelper(curIdx << 1, leftIdx, middle, idx, val); } else { this->upd…

题目 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…

题目如下: 解题思路:就三个字-线段树.这个题目是线段树用法最经典的场景. 代码如下: class NumArray(object): def __init__(self, nums): """ :type nums: List[int] """ self.nl = nums self.tree = [] if len(nums) == 0: return for i in xrange((4*len(nums)+1)): self.tree.a…

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…

问题描述:给定一序列,求任意区间(i, j)的元素和:修改任意一元素,实现快速更新 树状数组 树状数组的主要特点是生成一棵树,树的高度为logN.每一层的高度为k,分布在这一层的序列元素索引的二进制表达有个共同的特点,就是最低二次幂为k. 子树间有很强的联系,即,给定一序列元素索引i,可以推知该元素所在节点C[i]的父节点C[p],p=i+2^k,其中k=i & (i ^ (i-1)),以及该元素所在子树的前一子树的根节点p=i-2^k. 子树间的联系可以很方便地用于求前n个元素的和,即找到所有…

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…

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…

https://leetcode.com/problems/range-sum-query-mutable/ 因为数组会变动,所以缓存机制受到了挑战...每次更新数组意味着缓存失效,这样一更新一查找的话相当于每次都重新计算了. 所以要设计一个更好的缓存机制,尽量降低更新带来的影响. 我选择分段缓存,就是把原数组的缓存分别放在多段缓存里,这样数组变动的时候只用更新一段缓存. 我选择分成log(n) 个段,并没有什么道理. 这样在查询rangeSum 的时候就复杂了,如果范围在同一个缓存段内就很好,…

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…

在下面的图中,求[0,0]和[1,1]区间元素和, 对于[0,0] start =0 ,end = 0, 查到root = 4 [0-1]的节点,mid = 0, 此时mid > end-1,表示要查找的范围完全落到了4的左边节点,只递归左边 对于[1,1] start =1 ,end = 1, 查到root = 4 [0-1]的节点,mid = 0, 此时mid <start,表示要查找的范围完全落到了4的右边节点,只递归左边 对于[1,5],start = 1,end = 5,查到root…

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. 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…