版权声明：希望各位多提意见多多互相交流哦～ https://blog.csdn.net/wait_for_taht_day5/article/details/83281299
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].
Solution in C++：
关键点：
- unique & k可以为0
思路：
- 开始思路有点错误，直接用set去重了，然后以
(*it) + k的形式去寻找是否在set中，发现这样当k = 0的时候会返回set大小；然后就打算暴力了，先sort一下，然后直接nums[j] - nums[i] == k判断，这里存在的bug就是不能处理unique的情况，所以我用set存key的形式处理的，看到discuss里面也有人直接循环去重，因为这里是排序好的，所以相同的数字都挨在一起。
int findPairs(vector<int>& nums, int k) {
if (nums.size() == 0)
return 0;
int result = 0;
sort(nums.begin(), nums.end());
set<int> sets;
for(int i = 0; i < nums.size(); ++i){
for(int j = i+1; j < nums.size(); ++j){
if (nums[j] - nums[i] == k){
set<int>::iterator it = sets.find(nums[i]);
if ( it == sets.end()){
++result;
sets.insert(nums[i]);
}
break;
}
}
}
return result;
}
小结
今天有点弱，有点困了，明天还要加把劲干事情，另外一个没输完的题放到明天吧。今天算是把set的用法学了一遍。