求出最多标记下标

You are given a 0-indexed integer array nums.

Initially, all of the indices are unmarked. You are allowed to make this operation any number of times:

• Pick two different unmarked indices i and j such that 2 * nums[i] <= nums[j], then mark i and j.

Return the maximum possible number of marked indices in nums using the above operation any number of times.

 

Example 1:

Input: nums = [3,5,2,4]
Output: 2
Explanation: In the first operation: pick i = 2 and j = 1, the operation is allowed because 2 * nums[2] <= nums[1]. Then mark index 2 and 1.
It can be shown that there's no other valid operation so the answer is 2.

Example 2:

Input: nums = [9,2,5,4]
Output: 4
Explanation: In the first operation: pick i = 3 and j = 0, the operation is allowed because 2 * nums[3] <= nums[0]. Then mark index 3 and 0.
In the second operation: pick i = 1 and j = 2, the operation is allowed because 2 * nums[1] <= nums[2]. Then mark index 1 and 2.
Since there is no other operation, the answer is 4.

Example 3:

Input: nums = [7,6,8]
Output: 0
Explanation: There is no valid operation to do, so the answer is 0.

 

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 109

 

// 思路：
// 1. 对数组进行排序。
// 2. 使用Java Stream生成一个包含所有有效对的流。
// 3. 对有效对的数量求和。

import java.util.Arrays;
import java.util.stream.IntStream;

public class Solution {
    public int maxMarkedIndices(int[] nums) {
        Arrays.sort(nums);  // 对数组进行排序
        boolean[] marked = new boolean[nums.length];  // 标记数组
        return (int) IntStream.range(0, nums.length)
            .filter(i -> !marked[i])
            .flatMap(i -> IntStream.range(i + 1, nums.length)
                .filter(j -> 2 * nums[i] <= nums[j] && !marked[j])
                .peek(j -> {
                    marked[i] = true;  // 标记i
                    marked[j] = true;  // 标记j
                }))
            .count() * 2;  // 返回结果
    }
}

此题解采用排序加贪心的方法来解决问题。首先对数组 nums 进行排序，这样更容易找到满足 2 * nums[i] <= nums[j] 的 i 和 j。由于数组已排序，对于较小的元素，我们需要寻找后半部分中首个满足条件的较大元素。算法从数组的中点开始向后遍历，尝试找到与前半部分的元素 i 形成合法对的元素 j。如果找到了，i 指针向前移动，表示我们标记了一对下标。由于每次操作标记两个下标，所以返回的是成功匹配的对数乘以 2。

O(n log n)

O(1)