find pairs sum of x

Author: Manjunath

Data-Structure/LinkedList/doublyLinkedList/findPairsSumOfX.js

@@ -0,0 +1,58 @@
+'use strict';
+class Node {
+    constructor(data) {
+        this.data = data;
+        this.next = this.previous = null;
+    }
+}
+
+function findPairsSumOfX(list, x) {
+    let count = 0;
+    let head = list,
+        tail = list.previous;
+    while (head != tail) {
+        let sum = head.data + tail.data;
+        if (sum == x) {
+            head = head.next;
+            tail = tail.previous;
+            count++;
+            continue;
+        } else if (sum > x) tail = tail.previous;
+        else if (sum < x) head = head.next;
+    }
+    return count;
+}
+
+function displayList(list) {
+    let output = ''
+    let temp = list;
+    while (temp.next != list) {
+        output = output + temp.data + ' ';
+        temp = temp.next;
+    }
+    output = output + temp.data + ' ';
+    console.log(output);
+}
+
+let list = new Node(1);
+list.next = new Node(2);
+list.next.previous = list;
+
+list.next.next = new Node(3);
+list.next.next.previous = list.next;
+
+list.next.next.next = new Node(4);
+list.next.next.next.previous = list.next.next;
+
+list.next.next.next.next = new Node(5);
+list.next.next.next.next.previous = list.next.next.next;
+
+list.next.next.next.next.next = new Node(6);
+list.next.next.next.next.next.previous = list.next.next.next.next;
+
+list.previous = list.next.next.next.next.next;
+list.next.next.next.next.next.next = list;
+displayList(list);
+
+let x = 5;
+console.log('Pairs of ' + x + ' = ' + findPairsSumOfX(list, x));