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95 lines
1.9 KiB
Markdown
95 lines
1.9 KiB
Markdown
# [532. K-diff Pairs in an Array](https://leetcode.com/problems/k-diff-pairs-in-an-array/)
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## 题目
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Given an array of integers and an integer k, you need to find the number of unique k-diff pairs in the array. Here a k-diff pair is defined as an integer pair (i, j), where i and j are both numbers in the array and their absolute difference is k.
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**Example 1**:
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```
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Input: [3, 1, 4, 1, 5], k = 2
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Output: 2
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Explanation: There are two 2-diff pairs in the array, (1, 3) and (3, 5).
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Although we have two 1s in the input, we should only return the number of unique pairs.
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```
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**Example 2**:
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```
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Input:[1, 2, 3, 4, 5], k = 1
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Output: 4
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Explanation: There are four 1-diff pairs in the array, (1, 2), (2, 3), (3, 4) and (4, 5).
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```
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**Example 3**:
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```
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Input: [1, 3, 1, 5, 4], k = 0
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Output: 1
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Explanation: There is one 0-diff pair in the array, (1, 1).
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```
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**Note**:
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1. The pairs (i, j) and (j, i) count as the same pair.
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2. The length of the array won't exceed 10,000.
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3. All the integers in the given input belong to the range: [-1e7, 1e7].
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## 题目大意
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给定一个数组，在数组里面找到几组不同的 pair 对，每个 pair 对相差 K 。问能找出多少组这样的 pair 对。
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## 解题思路
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这一题可以用 map 记录每个数字出现的次数。重复的数字也会因为唯一的 key，不用担心某个数字会判断多次。遍历一次 map，每个数字都加上 K 以后，判断字典里面是否存在，如果存在， count ++，如果 K = 0 的情况需要单独判断，如果字典中这个元素频次大于 1，count 也需要 ++。
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## 代码
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```go
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package leetcode
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func findPairs(nums []int, k int) int {
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if k < 0 || len(nums) == 0 {
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return 0
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}
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var count int
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m := make(map[int]int, len(nums))
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for _, value := range nums {
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m[value]++
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}
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for key := range m {
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if k == 0 && m[key] > 1 {
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count++
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continue
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}
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if k > 0 && m[key+k] > 0 {
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count++
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}
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}
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return count
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}
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``` |