题目

You are given an integer array nums and an integer k.

In one operation, you can pick two numbers from the array whose sum equals k and remove them from the array.

Return the maximum number of operations you can perform on the array.

Example 1:

  1. Input: nums = [1,2,3,4], k = 5
  2. Output: 2
  3. Explanation: Starting with nums = [1,2,3,4]:
  4. - Remove numbers 1 and 4, then nums = [2,3]
  5. - Remove numbers 2 and 3, then nums = []
  6. There are no more pairs that sum up to 5, hence a total of 2 operations.

Example 2:

  1. Input: nums = [3,1,3,4,3], k = 6
  2. Output: 1
  3. Explanation: Starting with nums = [3,1,3,4,3]:
  4. - Remove the first two 3's, then nums = [1,4,3]
  5. There are no more pairs that sum up to 6, hence a total of 1 operation.

Constraints:

  • 1 <= nums.length <= 10^5
  • 1 <= nums[i] <= 10^9
  • 1 <= k <= 10^9

题意

在数组中找到一对数,使它们的和为指定值,求这样的对数的最大值。

思路

对数组排序后,直接two pointers求和即可。


代码实现

Java

  1. class Solution {
  2. public int maxOperations(int[] nums, int k) {
  3. int count = 0;
  4. int left = 0, right = nums.length - 1;
  5. Arrays.sort(nums);
  6. while (left < right) {
  7. int sum = nums[left] + nums[right];
  8. if (sum > k) {
  9. right--;
  10. } else if (sum < k) {
  11. left++;
  12. } else {
  13. count++;
  14. left++;
  15. right--;
  16. }
  17. }
  18. return count;
  19. }
  20. }