}voidupdateFromSon(intcurIdx){intleftIdx = curIdx <<1;intrightIdx = leftIdx |1;this->nodes[curIdx].sum =this->nodes[leftIdx].sum +this->nodes[rightIdx].sum; }intsumHelper(intcurIdx,intleftIdx,intrightIdx,intleftRange,intrightRange){// 不在范围内if(leftIdx > rightRange || right...
看了一下solution,神奇的发现用一个长度为nums.length+1的数组就可以避免两种特殊情况的讨论了。代码如下: 1privateint[] sum;23publicNumArray(int[] nums) {4sum =newint[nums.length + 1];5for(inti = 0; i < nums.length; i++) {6sum[i + 1] = sum[i] +nums[i];7}8}910publicintsumRa...
intsumIK(inti,intk)//i~k的区间和{ sum(i); sum(k+1);returnsum(k+1)-sum(i);//借用前面的求和,但是要注意,实际上应该是k-(i-1),因为从1开始,所以两边都要加上1} Solution: publicclassNumArray {int[] btree;int[] arr;publicNumArray(int[] nums) { btree=newint[nums.length+1]; arr...
1. Description Range Sum Query - Immutable 2. Solution Version 1 class NumArray{public:NumArray(vector<int>nums){this->nums=nums;}intsumRange(inti,intj){intsum=0;for(intindex=i;index<=j;index++){sum+=nums[index];}returnsum;}private:vector<int>nums;};/** * Your NumArray object wil...
sumRange(i, j) = sumRange(0, j) - sumRange(0, i - 1); 代码:C++ classNumArray {public: NumArray(vector<int> &nums) { num=nums;if(num.size() >0) { solution.push_back(num[0]);for(inti =1; i <= num.size() -1; i++) {intindex = solution[i -1] +num[i]; ...
https://leetcode.com/problems/range-sum-query-immutable/discuss/75184/5-lines-C%2B%2B-4-lines-Pythonhttps://leetcode.com/problems/range-sum-query-immutable/discuss/75192/Java-simple-O(n)-init-and-O(1)-query-solutionLeetCode All in One 题目讲解汇总(持续更新中...)...
You may assume the number of calls toupdateandsumRangefunction is distributed evenly. 这道题是之前那道Range Sum Query - Immutable的延伸,之前那道题由于数组的内容不会改变,所以我们只需要建立一个累计数组就可以支持快速的计算区间值了,而这道题说数组的内容会改变,如果我们还是用之前的方法建立累计和数组,...
建议和leetcode 307. Range Sum Query - Mutable 树状数组的一个应用 一起学习 代码如下: /* * 有使用归并排序或者线段树的该方法,可以优化到 O(long(n)) * 不过我肯定是想不到 * 所以我这里的是循环遍历的 O(n^2)解法 * */ class Solution ...
LeetCode-Range Sum Query - Mutable Given an integer arraynums, find the sum of the elements between indicesiandj(i≤j), inclusive. Theupdate(i, val)function modifiesnumsby updating the element at indexitoval. Example: Given nums = [1, 3, 5]...
* LeetCode 307. Range Sum Query - Mutable * 区间求和,单点修改问题 * 其解决方法主要为树状数组和线段树 * * 此代码使用树状数组 * * 两点需要注意: * 1.树状数组下标从 1 开始,题目中数据数组从 0 开始 * 解决:index 整体 +1 即可 * 2.树状数组支持单点增加,此题为单点修改。