题目描述

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 than that.

Example:

Input: nums = [-2,5,-1], lower = -2, upper = 2,
Output: 3
Explanation: The three ranges are : [0,0], [2,2], [0,2] and their respective sums are: -2, -1, 2.

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/count-of-range-sum

解题思路

我自己其实是没有什么思路的，只想到了暴力加前缀和，其实也思考了线段树，但是线段树求解区间和的速度比前缀和明显慢很多。

看了题解的第一种解法，并自己实现了一下，爆了一发long long，第二遍就过了，虽然思路不是自己的但是还是很开心的。

决定跟随官方题解仔细研究一下这道题。

题解一

主要的思路在于考虑解决一个子问题：两个升序排列的数组$a,b$，求$a[i]<b[j]$的$(i,j)$对数量

实现代码：

class Solution {
public:typedef long long ll;ll  ans = 0;ll  Lower, Upper;void Merge(vector<ll> &sum, vector<ll> &tmp, int l, int r){if(r - l <= 1) return ;int mid = (l + r) >> 1;Merge(sum, tmp, l, mid);Merge(sum, tmp, mid, r);int pl = mid, pr = mid;int idx = l;while(idx < mid){while(pl < r && sum[pl] < sum[idx] + Lower) ++pl;while(pr < r && sum[pr] <= sum[idx] + Upper) ++pr;ans += pr - pl;++idx;}int  i = l, j = mid;idx = l;while(i < mid || j < r){if(j >= r || i < mid && sum[i] < sum[j]){tmp[idx++] = sum[i++];} else {tmp[idx++] = sum[j++];}}for(int i=l; i<r; ++i){sum[i] = tmp[i];}}int countRangeSum(vector<int>& nums, int lower, int upper) {int n = nums.size();Lower = lower; Upper = upper;vector<long long > sum(n+1, 0);vector<long long > tmp(n+1, 0);for(int i=1; i<=n; ++i){sum[i] = sum[i-1] + nums[i-1];}Merge(sum, tmp, 0, n+1);return ans;}
};