solve problem K Diff Pairs In An Array

Author: P-ppc

README.md

@@ -187,6 +187,7 @@ All solutions will be accepted!
 |7|[Reverse Integer](https://leetcode-cn.com/problems/reverse-integer/description/)|[java/py/js](./algorithms/ReverseInteger)|Easy|
 |507|[Perfect Number](https://leetcode-cn.com/problems/perfect-number/description/)|[java/py/js](./algorithms/PerfectNumber)|Easy|
 |840|[Magic Squares In Grid](https://leetcode-cn.com/problems/magic-squares-in-grid/description/)|[java/py/js](./algorithms/MagicSquaresInGrid)|Easy|
+|532|[K Diff Pairs In An Array](https://leetcode-cn.com/problems/k-diff-pairs-in-an-array/description/)|[java/py/js](./algorithms/KDiffPairsInAnArray)|Easy|
 
 # Database
 |#|Title|Solution|Difficulty|

algorithms/KDiffPairsInAnArray/README.md

@@ -0,0 +1,2 @@
+# K Diff Pairs In An Array
+This problem is easy to solve by double pointers or hashmap

algorithms/KDiffPairsInAnArray/Solution.java

@@ -0,0 +1,30 @@
+class Solution {
+    public int findPairs(int[] nums, int k) {
+        if (k < 0) return 0;
+        
+        int res = 0,
+            length = nums.length,
+            i = 0,
+            j = 0;
+        
+        Arrays.sort(nums);
+        
+        while (i < length && j < length) {
+            if (i >= j) {
+                j++;
+            } else if (nums[j] - nums[i] == k) {
+                res++;
+                int temp = nums[i];
+                while (i < length && temp == nums[i]) {
+                    i++;
+                }
+            } else if (nums[j] - nums[i] > k) {
+                i++;
+            } else if (nums[j] - nums[i] < k) {
+                j++;
+            }
+        }
+        
+        return res;
+    }
+}

algorithms/KDiffPairsInAnArray/solution.js

@@ -0,0 +1,37 @@
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number}
+ */
+var findPairs = function(nums, k) {
+    if (k < 0) return 0
+    
+    let parisMap = {},
+        valueMap = {},
+        res = 0
+    
+    nums.forEach(num => {
+        let p1 = num + k,
+            p2 = num - k
+        
+        if (valueMap[p1]) {
+            let max = Math.max(p1, num),
+                min = Math.min(p1, num)
+            if (!parisMap[`${max}x${min}`]) {
+                parisMap[`${max}x${min}`] = true
+                res++
+            }
+        }
+        if (valueMap[p2]) {
+            let max = Math.max(p2, num),
+                min = Math.min(p2, num)
+            if (!parisMap[`${max}x${min}`]) {
+                parisMap[`${max}x${min}`] = true
+                res++
+            }
+        }
+        valueMap[num] = true
+    })
+    
+    return res
+};

algorithms/KDiffPairsInAnArray/solution.py

@@ -0,0 +1,26 @@
+class Solution(object):
+    def findPairs(self, nums, k):
+        """
+        :type nums: List[int]
+        :type k: int
+        :rtype: int
+        """
+        res = 0
+        length = len(nums)
+        nums.sort()
+        
+        i = j = 0
+        while j < length and i < length:
+            if i >= j:
+                j += 1
+            elif nums[j] - nums[i] == k:
+                res += 1
+                temp = nums[i]
+                while i < length and temp == nums[i]:
+                    i += 1
+            elif nums[j] - nums[i] > k:
+                i += 1
+            elif nums[j] - nums[i] < k:
+                j += 1
+            
+        return res