64 lines
1.8 KiB
Java
64 lines
1.8 KiB
Java
//小力将 N 个零件的报价存于数组 `nums`。小力预算为 `target`，假定小力仅购买两个零件，要求购买零件的花费不超过预算，请问他有多少种采购方案。
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//
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//
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//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`
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//
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//
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//**示例 1：**
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//>输入：`nums = [2,5,3,5], target = 6`
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//>
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//>输出：`1`
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//>
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//>解释：预算内仅能购买 nums[0] 与 nums[2]。
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//
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//**示例 2：**
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//>输入：`nums = [2,2,1,9], target = 10`
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//>
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//>输出：`4`
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//>
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//>解释：符合预算的采购方案如下：
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//>nums[0] + nums[1] = 4
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//>nums[0] + nums[2] = 3
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//>nums[1] + nums[2] = 3
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//>nums[2] + nums[3] = 10
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//
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//**提示：**
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//- `2 <= nums.length <= 10^5`
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//- `1 <= nums[i], target <= 10^5`
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// 👍 25 👎 0
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package leetcode.editor.cn;
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//LCP 28:采购方案
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public class FourXy4Wx{
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public static void main(String[] args) {
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//测试代码
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Solution solution = new FourXy4Wx().new Solution();
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}
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//力扣代码
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//leetcode submit region begin(Prohibit modification and deletion)
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class Solution {
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public int purchasePlans(int[] nums, int target) {
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int[] sort = new int[target];
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long[] count = new long[target];
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for (int num : nums) {
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if (num < target) {
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sort[num] += 1;
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}
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}
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long sum = 0;
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for (int i = 1; i < target; i++) {
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sum += sort[i];
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count[i] = sum;
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}
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long result = 0;
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for (int num : nums) {
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if (target > num) {
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result += num <= target - num ? count[target - num] - 1 : count[target - num];
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}
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}
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return (int) (result / 2 % (Math.pow(10, 9) + 7));
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}
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}
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//leetcode submit region end(Prohibit modification and deletion)
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} |