<LeetCode OJ> 327. Count of Range Sum

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本文介绍了一个算法问题——计数范围内和，即计算数组中所有子数组的和落在指定区间内的总数。文章首先给出了一个时间复杂度为O(n^2)的简单解法，并通过实例解释了如何计算每个子数组的和。随后，文章提出了一种更高效的解法，利用分治思想和归并排序中的思想来优化计算过程。

327. Count of Range Sum

My Submissions

Total Accepted: 93 Total Submissions: 316 Difficulty: Hard

给定一个整数数组nums，返回该数组某个"范围内的和"在 [ lower,upper ] 之间的总数量

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.注意：你的算法必须比O(N^2)性能更快

Example:
Given nums = [-2, 5, -1], lower = -2, upper = 2,
Return 3.
The three ranges are : [0, 0], [2, 2], [0, 2] and their respective sums are: -2, -1, 2.

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

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分析：

1，朴素的模拟思路：

1），令S(i)表示前i个数之和，比如，比如【-2，5，-1】，s(0)=-2,s(1)=3，s(2)=2...先将所有的和计算出来

2），令S(i, j)表示数组中位置i到j的元素之和，S(i, j)=S(j)-S(i)+a(i)

但是这样做time,o(n^2),space,o(n), Time Limit Exceeded

class Solution {
public:
    int countRangeSum(vector<int>& nums, int lower, int upper) {
        if(nums.size()==0)
            return 0;
        vector<int> sums(nums.size(),0);
        sums[0]=nums[0];
        for (int i = 1; i < nums.size(); ++i)
            sums[i] = sums[i-1] + nums[i];
        int cnt = 0;
        for (int i = 0; i < nums.size(); ++i)
            for (int j = i; j < nums.size(); ++j)
            {
                int gapsum=sums[j] - sums[i]+nums[i];
                if ( gapsum>= lower && gapsum <= upper)
                    cnt++;
            }
        return cnt;
    }
};

优化中.......若干时间后....不会！

参考讨论区：对两层for循环进行优化

class Solution {
public:
    int countRangeSum(vector<int>& nums, int lower, int upper) {
        int size_nums=nums.size();
        vector<long> sums(size_nums+1, 0);
        for(int i=0; i<size_nums; i++)   
            sums[i+1]=sums[i]+nums[i];
        return help(sums, 0, size_nums+1, lower, upper);
    }
    
    int help(vector<long> &sums, int start, int end, int lower, int upper){
        /***  end condition  ***/
        if(end - start <= 1) return 0;
        /***  recursive devidely  ***/
        int mid=(start+end)/2;
        int count=help(sums, start, mid, lower, upper) + help(sums, mid, end, lower, upper);
        int j=mid, k=mid, t=mid;
        vector<long> cache(end-start, 0);
        for(int i=start, r=0; i<mid; i++, r++)
        {
            while(k<end && sums[k]-sums[i] < lower) k++;
            while(j<end && sums[j]-sums[i] <= upper) j++;
            /***  merge sorting parts  ***/
            while(t<end && sums[t] < sums[i])  cache[r++]=sums[t++];
            cache[r]=sums[i];
            count+=j-k;
        }
        //copy the merge sorted array to the original sums
        for(int i=0; i<t-start; i++)  
            sums[start+i]=cache[i];
        return count;
    }
};

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原文地址：http://blog.youkuaiyun.com/ebowtang/article/details/50492602

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