小扣在秋日市集选择了一家早餐摊位,一维整型数组 staple 中记录了每种主食的价格,一维整型数组 drinks 中记录了每种饮料的价格。小扣的计划选择一份主食和一款饮料,且花费不超过 x 元。请返回小扣共有多少种购买方案。
注意:答案需要以 1e9 + 7 (1000000007) 为底取模,如:计算初始结果为:1000000008,请返回 1
示例 1:
输入:staple = [10,20,5], drinks = [5,5,2], x = 15
输出:6
解释:小扣有 6 种购买方案,所选主食与所选饮料在数组中对应的下标分别是:
第 1 种方案:staple[0] + drinks[0] = 10 + 5 = 15;
第 2 种方案:staple[0] + drinks[1] = 10 + 5 = 15;
第 3 种方案:staple[0] + drinks[2] = 10 + 2 = 12;
第 4 种方案:staple[2] + drinks[0] = 5 + 5 = 10;
第 5 种方案:staple[2] + drinks[1] = 5 + 5 = 10;
第 6 种方案:staple[2] + drinks[2] = 5 + 2 = 7。
示例 2:
输入:staple = [2,1,1], drinks = [8,9,5,1], x = 9
输出:8
解释:小扣有 8 种购买方案,所选主食与所选饮料在数组中对应的下标分别是:
第 1 种方案:staple[0] + drinks[2] = 2 + 5 = 7;
第 2 种方案:staple[0] + drinks[3] = 2 + 1 = 3;
第 3 种方案:staple[1] + drinks[0] = 1 + 8 = 9;
第 4 种方案:staple[1] + drinks[2] = 1 + 5 = 6;
第 5 种方案:staple[1] + drinks[3] = 1 + 1 = 2;
第 6 种方案:staple[2] + drinks[0] = 1 + 8 = 9;
第 7 种方案:staple[2] + drinks[2] = 1 + 5 = 6;
第 8 种方案:staple[2] + drinks[3] = 1 + 1 = 2;
提示:
- 1 <= staple.length <= 10^5
- 1 <= drinks.length <= 10^5
- 1 <= staple[i],drinks[i] <= 10^5
- 1 <= x <= 2*10^5
Code:
class Solution {
public:
int breakfastNumber(vector<int>& staple, vector<int>& drinks, int x) {
int count=0;
map<int,int>mymap1;
for(int i=0;i<staple.size();i++)
{
mymap1[staple[i]]++;
}
map<int,int>mymap2;
for(int i=0;i<drinks.size();i++)
{
mymap2[drinks[i]]++;
}
map<int,int>::iterator it;
map<int,int>::iterator it2;
for(it=mymap1.begin();it!=mymap1.end();it++)
{
for(it2=mymap2.begin();it2!=mymap2.end();it2++)
{
if(((it->first)+(it2->first))<=x)
{
count+=(it->second*it2->second);
}
}
}
return count%1000000007;
}
};
网友思路:
- 排序
- 双指针
- i 从前往后, j 从后往前,找到 staple[i] + drinks[j] <= x 的位置 就可以往答案里加了
class Solution {
public:
int breakfastNumber(vector<int>& staple, vector<int>& drinks, int x) {
const int mod = 1e9 + 7;
int ans = 0;
sort(staple.begin(), staple.end());
sort(drinks.begin(), drinks.end());
int j = drinks.size() - 1;
for (int i = 0; i < staple.size(); i++) {
while (j >= 0 && staple[i] + drinks[j] > x) j--;
if (j == -1) break;
ans += j + 1;
ans %= mod;
}
return ans;
}
};