LeetCode - 532 - K-diff Pairs in an Array

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].

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题意：找出差值为k的整数对有多少（去重）

思路：扫一遍存map里，再扫一遍做判断。注意k==0的情况。

最开始写了个O(nlogn)的，排个序二分找，后来看别人博客发现可以O(n)，借助了个map时间复杂度瞬间降下来了。果然自己还是很弱呢，加油啦~

男票说我得学学C++的写法，叹气，果然总是写C的代码不行呢【托腮】

class Solution {
public:
int findPairs(vector<int>& nums, int k) {
int len = nums.size();
map<int, int> m;
m.clear();
int ans = 0;
for (int i = 0; i < len; i++) {
m[nums[i]]++;
}
for (auto a: m) { //这是刚刚接触的比较神奇的东西，C++11的新特性
if (!k && a.second > 1) ans++;
if (k && m.count(a.first + k)) ans++;
}
return ans;
}
};