6.19 removeNthNodeFromEnd &

• 6.19
• to do

1] Remove Nth Node From End of List

unhandled special cases arise w/o help of dummy

    ListNode* removeNthFromEnd(ListNode* head, int n) {
        int len = 0;
        for (ListNode* curr=head; curr; curr=curr->next) {
            ++len;
        }

        ListNode dummy(-1);
        ListNode* finder = &dummy;
        dummy.next = head;
        for (int steps=0; steps<len-n; ++steps) {
            finder = finder->next;
        }
        ListNode* remv = finder->next;
        finder->next = remv->next;
        delete(remv);
        return dummy.next;
    }

2] Swap Nodes in Pairs

    ListNode* swapPairs(ListNode* head) {
        if (!head) return head;
        ListNode dummy(-1);
        dummy.next = head;

        for ( ListNode* prev=&dummy, *l=prev->next, *r=l->next;
              r;prev=l, l=prev->next, r=!l? nullptr : l->next) {
            prev->next = r;
            l->next = r->next;
            r->next = l;
        }
        return dummy.next;
    }

note： in for loop, r=!l? nullptr : l->next

3] Reverse Nodes in k-Group

== redo w/o helper, try recursion ==

note while (--n>0) {} => body executes for n-1 times

    ListNode* reverseList(ListNode* head) {
        if (!head || !head->next) return head;
        ListNode* newh = reverseList(head->next);
        head->next->next = head;
        head->next = NULL;
        return newh;
    }

    ListNode* reverseKGroup(ListNode* head, int k) {
        ListNode dummy(-1);
        dummy.next=head;
        int ct=0;
        for (ListNode* prev=&dummy, *curr=prev->next; curr; curr = prev->next) {
            int steps = k; //do k-1 times
            while (curr && --steps>0) { curr = curr->next; }
            if (!curr) break;

            ListNode* nextGroup = curr->next;
            curr->next = NULL;
            ListNode* oldHead = prev->next;
            prev->next = reverseList(oldHead);
            oldHead->next = nextGroup;
            prev = oldHead;

        }
        return dummy.next;
    }

4] Copy List with Random Pointer
51% okiee :3

== redo w/ 分拆链表 ==

    RandomListNode *copyRandomList(RandomListNode *head) {
        RandomListNode dummy(-1);
        dummy.next = head;
        RandomListNode* newh = &dummy;

        vector<RandomListNode*> newAddr;
        unordered_map<RandomListNode*, int> addressToIndex;

        // writer always points to the elem last written
        int ct = 0;
        for (RandomListNode* reader=head, *writer=newh; reader; reader=reader->next, writer=writer->next) {
            writer->next = new RandomListNode(reader->label);
            addressToIndex[reader] = (ct++);
            newAddr.push_back(writer->next);
        }

        for (RandomListNode* reader=head, *writer=newh; reader; reader=reader->next, writer=writer->next) {
            if (reader->random) {
                writer->next->random = newAddr[ addressToIndex[reader->random] ];
            }
        }
        return dummy.next;

    }

- btw

5] Rectangle Area
or note if there's overlap, its area = [min(rightXcoords)-max(leftXcoords)] * [min(higherYcoords)-max(lowerYcoords)]

    // make sure input x1l is the leftmost point
    int overlapLen(int x1l, int x1r, int x2l, int x2r) {
        if (x2l>=x1r) return 0;
        if (x2r<=x1r) return x2r-x2l;
        else return x1r-x2l;
    }

    int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
        int xlen = A<=E? overlapLen(A, C, E, G) : overlapLen(E, G, A, C);
        int ylen = B<=F? overlapLen(B, D, F, H) : overlapLen(F, H, B, D);
        return (C-A)*(D-B) + (G-E)*(H-F) - (xlen*ylen);
    }