532. K-diff Pairs in an Array

原题

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].

代码分析

public int FindPairs(int[] nums, int k)
{
int sum = 0;
if (nums.Length == 0 || nums.Length == 1) return 0;
System.Array.Sort(nums);
for (int i = 0; i < nums.Length; i++)
{
bool eql = false;
for (int j = i + 1; j < nums.Length; j++)
{
while (j < nums.Length && nums[j] == nums[i])
{
j++;
eql = true;

4000
}
if (eql) //找到了相等元素
{
i = j - 1; //j-1>i=0 //i等于某一块相等元素的末端点
if (k == 0)
{
sum++;
break;
}
eql = false;
}
while (j + 1 < nums.Length && nums[j + 1] == nums[j])
j++; //j等于后一块相等元素的末端点
if (j < nums.Length && nums[j] - nums[i] > k) //
break;
if (j < nums.Length && nums[j] - nums[i] == k)
sum++;
}
}
return sum;
}

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