Linked Lists: Medium

Add two numbers

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode result = new ListNode(-1);
        ListNode rcur = result;
        ListNode cur1 = l1;
        ListNode cur2 = l2; 
        int carry = 0; 
        while (cur1 != null && cur2 != null) {
            int cur = cur1.val + cur2.val + carry;
            rcur.next = new ListNode(cur % 10);
            
            carry = cur >= 10 ? cur / 10 : 0;
            
            cur1 = cur1.next;
            cur2 = cur2.next; 
            rcur = rcur.next;
        }
        
        if (cur1 != null) {
            rcur.next = cur1;
        } else {
            rcur.next = cur2; 
        }
        
        return result.next; 
    }
}

Wrong answer
digit수가 다를 경우를 생각 X

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode result = new ListNode(-1);
        ListNode rcur = result;
        ListNode cur1 = l1;
        ListNode cur2 = l2; 
        int carry = 0; 
        while (cur1 != null && cur2 != null) {
            int cur = cur1.val + cur2.val + carry;
            rcur.next = new ListNode(cur % 10);
            
            carry = cur >= 10 ? cur / 10 : 0;
            
            cur1 = cur1.next;
            cur2 = cur2.next; 
            rcur = rcur.next;
        }
        
        while (cur1 != null) {
            int c1 = cur1.val + carry;
            rcur.next = new ListNode(c1 % 10);
            carry = c1 >= 10 ? c1 / 10 : 0;
            cur1 = cur1.next;
            rcur = rcur.next; 
        }
        
        while (cur2 != null) {
            int c2 = cur2.val + carry;
            rcur.next = new ListNode(c2 % 10);
            carry = c2 >= 10 ? c2 / 10 : 0;
            cur2 = cur2.next;
            rcur = rcur.next; 
        }
        
        if (carry != 0) {
            rcur.next = new ListNode(carry);
        }
        
        return result.next; 
    }
}

혼자서 풀어서 너무 뿌듯!!
저번에 풀었을때에는 루션이를 봤었네요^^..

같은 아이디어인데 훨씬 깔끔하게 쓰여짐

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode dummyHead = new ListNode(0);
    ListNode p = l1, q = l2, curr = dummyHead;
    int carry = 0;
    while (p != null || q != null) {
        int x = (p != null) ? p.val : 0;
        int y = (q != null) ? q.val : 0;
        int sum = carry + x + y;
        carry = sum / 10;
        curr.next = new ListNode(sum % 10);
        curr = curr.next;
        if (p != null) p = p.next;
        if (q != null) q = q.next;
    }
    if (carry > 0) {
        curr.next = new ListNode(carry);
    }
    return dummyHead.next;
}
}

Odd Even Linked List

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode oddEvenList(ListNode head) {
        ListNode evens = new ListNode(-1);
        ListNode odds = new ListNode(-1);
        ListNode oc = odds;
        ListNode ec = evens;
        ListNode cur = head;
        boolean isOdd = true; 
        
        while (cur != null) {
            if (isOdd) {
                odds.next = cur;
                odds = odds.next;
            } else {
                evens.next = cur;
                evens = evens.next;
            }
            isOdd = ! isOdd;
            cur = cur.next; 
            
        }
        evens.next = null;
        odds.next = ec.next;
        
        return oc.next;
        
    }
}

뿌듯뿌듯~~

이거 처음에 풀었을떄는 파이선으로 풀었었네요;
지금은 파이선 하나도 못하겠다는점^_^

Intersection of Two Linked Lists

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        if (headA == null || headB == null) {
            return null;
        }
        
        ListNode cura = headA;
        ListNode curb = headB;
        
        while (cura != curb) {
            cura = cura == null ? headB : cura.next;
            curb = curb == null? headA : curb.next;
        }
        
        return cura;
        
    }
}

아직도 충격적인 코드..
냅다 외운게 효과가 있네요^__^