//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` Related Topics 数组 双指针 二分查找 排序 // 👍 50 👎 0 package leetcode.editor.cn; //LCP 18:早餐组合 public class TwoVYnGI { public static void main(String[] args) { //测试代码 Solution solution = new TwoVYnGI().new Solution(); } //力扣代码 //leetcode submit region begin(Prohibit modification and deletion) class Solution { public int breakfastNumber(int[] staple, int[] drinks, int x) { int[] array = new int[x + 1]; for (int num : staple) { if (num < x) { array[num]++; } } for (int i = 1; i < x + 1; i++) { array[i] += array[i - 1]; } int count = 0; for (int num : drinks) { if (num < x) { count += array[x - num]; count %= 1000000007; } } return count; } } //leetcode submit region end(Prohibit modification and deletion) }