1+ /*
2+ Return the 'middle' node of a linked list.
3+ If the list has an even number of elements, return the node at the end of the first half of the list.
4+
5+ *Do not* use a counter variable, *do not* retrive the size of the list, and only iterate through the list one time.
6+
7+ --Example
8+
9+ const l = new LinkedList();
10+ l.insertLast('a');
11+ l.insertLast('b');
12+ l.insertLast('c');
13+
14+ midPoint(l); // returns { data: 'b' };
15+ */
16+ 17+ // -----------------------------------------------
18+ 19+ const oddArr = [ 4 , 7 , 5 , 9 , 2 , 47 , 11 ] ; // 9
20+ const evenArr = [ 4 , 7 , 5 , 9 , 2 , 47 ] // 5
21+ 22+ function midPoint ( arr ) {
23+ let slow = 0 ;
24+ let fast = 0 ;
25+ 26+ while ( arr [ fast + 2 ] ) {
27+ slow += 1 ;
28+ fast += 2 ;
29+ }
30+ 31+ return arr [ slow ] ;
32+ }
33+ 34+ console . log ( midPoint ( oddArr ) ) ; // 9
35+ console . log ( midPoint ( evenArr ) ) ; // 5
36+ 37+ // -----------------------------------------------
38+ 39+ // Solution technique:
40+ // We will use 2 pointer variables 'slow' and 'fast'.
41+ // With each iteration slow will move ahead by 1 and fast will move ahead by 2.
42+ // 'fast' will move twice as fast as 'slow'.
43+ // So, when 'fast' will point to the last node, 'slow' must be pointing to the middle node.
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