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532.K-diff Pairs in an Array
532.K-diff Pairs in an Array
难度:Easy
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:
The pairs (i, j) and (j, i) count as the same pair.
The length of the array won't exceed 10,000.
All the integers in the given input belong to the range: [-1e7, 1e7].
考虑到k有可能为0，所以需要使用hash表。
class Solution {
public:
int findPairs(vector<int>& nums, int k) {
if(k<0) return 0;
unordered_map<int,int>N;
for(auto n:nums)
N[n]++;
int res=0;
if(k)
for(auto n:N)
{
if(N.count(n.first+k)) res++;
if(N.count(n.first -k)) res++;
}
else
{
for(auto n:N)
if(n.second >1) res+=2;
}
return res/2;
}
};
执行用时 :40 ms, 在所有 C++ 提交中击败了81.60%的用户 内存消耗 :12.1 MB, 在所有 C++ 提交中击败了40.60%的用户
Last modified 1yr ago