Leetcode Count of Range Sum

最新推荐文章于 2022-12-02 15:35:02 发布

markpen

最新推荐文章于 2022-12-02 15:35:02 发布

阅读量286

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode 分治; 文章标签：归并排序

本文链接：https://blog.youkuaiyun.com/markpen/article/details/67634410

leetcode 同时被 2 个专栏收录

167 篇文章

订阅专栏

分治;

2 篇文章

订阅专栏

题意：求连续和在区间内的和的个数。

思路：分治，利用归并排序的思想，分成前后两部分。如果sumb [L] - suma[I] > upper sumb[K] - suma[I] > lower，那么L - K之间的和都满足要求。

class Solution {
public:
    int countRangeSum(vector<int>& nums, int lower, int upper) {
        vector<double> sum;
        double tempSum = 0;
        sum.push_back(tempSum);
        for(int i = 0; i < nums.size(); ++ i) {
            tempSum += nums[i];
            sum.push_back(tempSum);
        }
        count = 0;
        up = upper, down = lower;
        
        //for(int i = 0; i < sum.size(); ++ i) cout << sum[i] << endl;
        sum = merge(sum);
        //for(int i = 0; i < sum.size(); ++ i) cout << sum[i] << endl;
        
        return count;
        
    }
    
    vector<double> merge(vector<double> nums) {
        if(nums.size() == 1) return nums;
        vector<double> first;
        vector<double> second;
        for(int i = 0; i < nums.size() / 2; ++ i) first.push_back(nums[i]);
        for(int i = nums.size() / 2; i < nums.size(); ++ i) second.push_back(nums[i]);
        first = merge(first); //cout << first.size();
        second = merge(second); //cout << second.size() << endl; cout << endl;
        if(first.size() == 0) return second;
        if(second.size() == 0) return first;
        
        vector<double> re;
        int j = 0;
        int k = 0, l = 0;
        for(int i = 0; i < first.size(); i ++ ) { //cout << second[k] - first[i] << endl;
            while(second[k] - first[i] < down && k < second.size()) k ++;
            while(second[l] - first[i] <= up && l < second.size()) l ++;
            while(first[i] > second[j] && j < second.size()) re.push_back(second[j ++]);
            re.push_back(first[i]);
            count += (l - k); //cout << l << " " << k << endl;
        }
        for(; j < second.size(); j ++) re.push_back(second[j]);
        return re;
        
    }
    
    int count;
    double up, down;
};