Leetcode Count of Range Sum

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;
};

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值