# [LeetCode] K-diff Pairs in an Array 数组中差为K的数对

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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.

Example 1:

Input: [3, 1, 4, 1, 5], k = 2
Output: 2
Explanation: There are two 2-diff pairs in the array, (1, 3) and (3, 5).
Although we have two 1s in the input, we should only return the number of unique pairs.


Example 2:

Input:[1, 2, 3, 4, 5], k = 1
Output: 4
Explanation: There are four 1-diff pairs in the array, (1, 2), (2, 3), (3, 4) and (4, 5).


Example 3:

Input: [1, 3, 1, 5, 4], k = 0
Output: 1
Explanation: There is one 0-diff pair in the array, (1, 1).


Note:

1. The pairs (i, j) and (j, i) count as the same pair.
2. The length of the array won't exceed 10,000.
3. All the integers in the given input belong to the range: [-1e7, 1e7].

class Solution {
public:
int findPairs(vector<int>& nums, int k) {
int res = 0, n = nums.size();
unordered_map<int, int> m;
for (int num : nums) ++m[num];
for (auto a : m) {
if (k == 0 && a.second > 1) ++res;
if (k > 0 && m.count(a.first + k)) ++res;
}
return res;
}
};

public:
int findPairs(vector<int>& nums, int k) {
int res = 0, n = nums.size(), j = 0;
sort(nums.begin(), nums.end());
for (int i = 0; i < n; ++i) {
int j = max(j, i + 1);
while (j < n && (long)nums[j] - nums[i] < k) ++j;
if (j < n && (long)nums[j] - nums[i] == k) ++res;
while (i < n - 1 && nums[i] == nums[i + 1]) ++i;
}
return res;
}
};

本文转自博客园Grandyang的博客，原文链接：K-diff Pairs in an Array 数组中差为K的数对，如需转载请自行联系原博主。

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