mirror of
https://github.com/halfrost/LeetCode-Go.git
synced 2025-07-11 22:20:34 +08:00
98 lines
2.5 KiB
Markdown
98 lines
2.5 KiB
Markdown
# [1679. Max Number of K-Sum Pairs](https://leetcode.com/problems/max-number-of-k-sum-pairs/)
|
||
|
||
|
||
## 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:**
|
||
|
||
```
|
||
Input: nums = [1,2,3,4], k = 5
|
||
Output: 2
|
||
Explanation: Starting with nums = [1,2,3,4]:
|
||
- Remove numbers 1 and 4, then nums = [2,3]
|
||
- Remove numbers 2 and 3, then nums = []
|
||
There are no more pairs that sum up to 5, hence a total of 2 operations.
|
||
```
|
||
|
||
**Example 2:**
|
||
|
||
```
|
||
Input: nums = [3,1,3,4,3], k = 6
|
||
Output: 1
|
||
Explanation: Starting with nums = [3,1,3,4,3]:
|
||
- Remove the first two 3's, then nums = [1,4,3]
|
||
There are no more pairs that sum up to 6, hence a total of 1 operation.
|
||
```
|
||
|
||
**Constraints:**
|
||
|
||
- `1 <= nums.length <= 105`
|
||
- `1 <= nums[i] <= 109`
|
||
- `1 <= k <= 109`
|
||
|
||
## 题目大意
|
||
|
||
给你一个整数数组 nums 和一个整数 k 。每一步操作中，你需要从数组中选出和为 k 的两个整数，并将它们移出数组。返回你可以对数组执行的最大操作数。
|
||
|
||
## 解题思路
|
||
|
||
- 读完题第一感觉这道题是 TWO SUM 题目的加强版。需要找到所有满足和是 k 的数对。先考虑能不能找到两个数都是 k/2 ，如果能找到多个这样的数，可以先移除他们。其次在利用 TWO SUM 的思路，找出和为 k 的数对。利用 TWO SUM 里面 map 的做法，时间复杂度 O(n)。
|
||
|
||
## 代码
|
||
|
||
```go
|
||
package leetcode
|
||
|
||
// 解法一 优化版
|
||
func maxOperations(nums []int, k int) int {
|
||
counter, res := make(map[int]int), 0
|
||
for _, n := range nums {
|
||
counter[n]++
|
||
}
|
||
if (k & 1) == 0 {
|
||
res += counter[k>>1] >> 1
|
||
// 能够由 2 个相同的数构成 k 的组合已经都排除出去了，剩下的一个单独的也不能组成 k 了
|
||
// 所以这里要把它的频次置为 0 。如果这里不置为 0，下面代码判断逻辑还需要考虑重复使用数字的情况
|
||
counter[k>>1] = 0
|
||
}
|
||
for num, freq := range counter {
|
||
if num <= k/2 {
|
||
remain := k - num
|
||
if counter[remain] < freq {
|
||
res += counter[remain]
|
||
} else {
|
||
res += freq
|
||
}
|
||
}
|
||
}
|
||
return res
|
||
}
|
||
|
||
// 解法二
|
||
func maxOperations_(nums []int, k int) int {
|
||
counter, res := make(map[int]int), 0
|
||
for _, num := range nums {
|
||
counter[num]++
|
||
remain := k - num
|
||
if num == remain {
|
||
if counter[num] >= 2 {
|
||
res++
|
||
counter[num] -= 2
|
||
}
|
||
} else {
|
||
if counter[remain] > 0 {
|
||
res++
|
||
counter[remain]--
|
||
counter[num]--
|
||
}
|
||
}
|
||
}
|
||
return res
|
||
}
|
||
``` |