Fifty-two face questions: the first point of intersection of the two lists

 

Method a: violence, traverse the linked list of each node A, the list B to be checked O (AB)
Method Two: Since the title list is unidirectional, so that after the first point of coincidence, the other are equal, that is Y-shaped. So they tail nodes are the same, and therefore weight behind the start line with "last-out characteristics", we can define two stacks, then the list into the stack, the stack during the comparison;
Method three: first get two lists of length, if the length of the longer length shorter than m, the step length m and then go down together and start comparing

Findfrist ListNode (ListNode A, B ListNode) {
             // Get the length of the list 
            int Alength =. 1, bLength =. 1 ; 
             ListNode A1 = A;
              the while (! A1.next = null ) { 
                    A1 = A1.next; 
                    Alength ++ ; 
             } 
             Bl ListNode = B;
              the while (B1.next =! null ) { 
                    Bl = B1.next; 
                    bLength ++ ; 
             } 
            // long go k steps
             int K = B.length- A.length;
              the while (K <0 ) { 
                K ++ ; 
                A = A.next; 
             } 
             the while (K> 0 ) { 
                K - ; 
                B = B.next; 
             } 
            // equivalent Comparative started until it finds the intersection of the direct return length · 
            the while (Alength> 0 ) {
                     IF (A == B) return A; 
                    A = A.next; 
                    B = B.next; 
                    A.length- ; 
            } 

            // no intersection in the end is null 
            return  null ; 
    }

Note: If the node as the root of the tail, then two trees looking like a common ancestor