Leetcode#532. K-diff Pairs in an Array(k差异对）

题目

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].

题意

给你一个数组，找出相差为k的不重复的数对，输入满足条件的数对的个数。

题解

只需判断当前数+k是否存在数组中，如果存在数组中则能构成一对数对。需要考虑去重差异对。

构建映射：数值 —>个数（C++使用map, python使用字典）

C++代码

class Solution {
public:
int findPairs(vector<int>& nums, int k) {

int sum = 0;
map<int,int>m;
if(k<0){ //差值为负数
return 0;
}
for(int i=0; i<nums.size(); i++){//构建数值-个数的映射关系
m[nums[i]]++;
}
map<int,int>::iterator it;

for(it=m.begin(); it!=m.end(); it++){//遍历map容器中的值，分两种情况讨论
if(k==0){//k==0且个数大于1的可以构成一个数对
if(it->second>1){
sum++;
}
}
else//k!=0,是如果当前值+k存在map中则可以构成一个数对
{
if(m.find(it->first + k) != m.end()){
sum++;
}
}

}
return sum;
}
};

python代码

class Solution(object):
def findPairs(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: int
"""
sum = 0
m = {}
if k<0:
return 0
for i in range(0, len(nums)):#初始化
m[nums[i]] = 0
for i in range(0, len(nums)):
m[nums[i]] = m[nums[i]] + 1
for key in m:
if k==0:
if m[key] > 1:
sum = sum + 1
else:
if m.has_key(key + k):
sum = sum + 1
return sum