LeetCode 532. K-diff Pairs in an Array

532. K-diff Pairs in an Array

一、问题描述

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.

二、输入输出

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.

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).

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

三、解题思路

注意是绝对距离，是

|i - j|

所以

k < 0

直接返回0。不要丢掉这种情况

假设k！=0 那么可以把数存到set中，set是集合，里面的元素不会重复。依次遍历，并查看

set(i) + k

是否也在set中：在，计数器加1；不在，继续遍历

对于

k=0

，同样借助set。在插入到set之前，首先判断是否已经包含该数字，如果有，说明出现了相同的数字，计数器加一。

注意

k=0

时，

<1,1>

这是1种，如果输入是

[1,1,1,1,1]

那么应该输出是1，不是4。所以我们需要把出现过的相同的数记下来，遍历过程中如果再次出现，直接忽略，不在考虑

int findPairs(vector<int>& nums, int k) {
set<int> s;
int n = nums.size(), ret = 0;
if( k < 0 ){
return 0;
}
else if( k == 0){
set<int> searched;
for (int i = 0; i < n; ++i) {
if (s.find(nums[i]) != s.end() && searched.count(nums[i]) == 0) {
ret++;
searched.insert(nums[i]);
}
else
s.insert(nums[i]);
}
}else{
for (int i = 0; i < n; ++i) {
s.insert(nums[i]);
}
for (set<int>::iterator ite = s.begin(); ite != s.end(); ite++){
if(s.find(*ite + k) != s.end())
ret++;
}
}
return ret;
}