Given an integer arraynums, return the number of range sums that lie in[lower, upper]inclusive. Range sumS(i, j)is defined as the sum of the elements innumsbetween indicesiandj(i≤j), inclusive. Note: A naive algorithm ofO(n2) is trivial. You MUST do better than that. Example: Give...
主要使用TreeMap的subMap的方法,求得落在区间内[sum[i] - high, sum[i] - lower]的sum[j]的个数。 public int countRangeSum(int[] nums, int lower, int upper) { if(nums == null || nums.length == 0){ return 0; } //key is the sum[i], value is the corresponding count // sum[...
Given an integer array nums, return the number of range sums that lie in [lower, upper] inclusive. Range sum S(i, j) is defined as the sum of the elements in nums between indices i and j (i ≤ j), inclusive. Note: A naive algorithm of O(n2) is trivial. You MUST do better t...
3、另外,如果元素sums[i]本身在范围内的话,需要在sums中加入元素0。 solution: classSolution(object):defcountRangeSum(self,nums,lower,upper):""" :type nums: List[int] :type lower: int :type upper: int :rtype: int """importbisect# 求前n项和nums,# 如果 lower <= nums[j] - nums[i] ...
LeetCode: 327. Count of Range Sum Given an integer array nums, return the number of range sums that lie in [lower, upper] inclusive. Range sum S(i, j) is defined as the sum of the elements in nums between indices i and j (i ≤ j), inclusive. ...
给定一个整数数组,返回range sum 落在给定区间[lower, upper] (包含lower和upper)的个数。range sum S(i, j) 表示数组中第i 个元素到j 个元素之和。 Note: A naive algorithm of O(n2) is trivial. You MUST do better than that. Example: ...
typedef long long int _int; class Solution { public: _int* l[10005*17]; _int* r[10005*17]; int sizel[10005*17]; int sizer[10005*17]; _int range[10005*17]; vector<int> num; _int low; _int upp; _int ans; int countRangeSum(vector<int>& nums, int lower, int upper) { if...
cache[r]=sums[i]; cnt+= j -k; } copy(cache.begin(), cache.begin()+ t - start, sums.begin() +start);returncnt; } }; 本文转自博客园Grandyang的博客,原文链接:区间和计数[LeetCode] Count of Range Sum,如需转载请自行联系原博主。
327. Count of Range Sum 327. Count of Range Sum 题目链接:https://leetcode.com/problems... 这题实际就是给定范围内的range sum,divide and conquer的方法。一路计算prefixSum[0:i],并把结果放进tree里面,然后计算到prefixSum[0:j+1]的时候,找tree里面有没有满足条件的prefixSum[0:i],这里的条件是...
// LeetCode #286 hard// 327. Count of Range Sum// Runtime: 112 ms, faster than 5.57% of C++ onlinesubmissionsfor Count of Range Sum.// Memory Usage: 9.2 MB, less than 100.00% of C++ online submissions for Count of Range Sum.classSolution{public:intcountRangeSum(vector<int>&nums,int...