LeetCode之链表

2. Add Two Numbers

You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

暴力解法：

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解题思路：

1.其中一个链表为空的情况；

2.链表长度相同时，且最后一个节点相加有进位的情况；

3.链表长度不等，短链表最后一位有进位的情况，且进位后长链表也有进位的情况；

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if(l1==null){
            return l2;
        }
        if(l2==null){
            return l1;
        }
        int len1 = getLength(l1);
        int len2 = getLength(l2);
        ListNode head = null;
        ListNode plus = null;
        if(len1>=len2){
            head = l1;
            plus = l2;
        }else{
            head = l2;
            plus = l1;
        }
        ListNode p = head;
        int carry = 0;
        int sum = 0;
        while(plus!=null){
            sum = p.val + plus.val+carry;
            carry = sum/10;
            p.val = sum%10;
            if(p.next==null&&carry!=0){
                ListNode node = new ListNode(carry);
                p.next = node;
                carry = 0;
            }
            p = p.next;
            plus = plus.next;
        }
        while(p!=null&&carry!=0){
            sum = p.val+carry;
            carry = sum/10;
            p.val = sum%10;
            if(p.next==null&&carry!=0){
                ListNode node = new ListNode(carry);
                p.next = node;
                carry=0;
            }
            p = p.next;
        }
        return head;
    }

    public int getLength(ListNode l){
        if(l==null){
            return 0;
        }
        int len = 0;
        while(l!=null){
            ++len;
            l = l.next;
        }
        return len;
    }
}

递归解法：

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复杂度

时间O(n) 空间(n) 递归栈空间

思路

从末尾到首位，对每一位对齐相加即可。技巧在于如何处理不同长度的数字，以及进位和最高位的判断。这里对于不同长度的数字，我们通过将较短的数字补0来保证每一位都能相加。递归写法的思路比较直接，即判断该轮递归中两个ListNode是否为null。

• 全部为null时，返回进位值
• 有一个为null时，返回不为null的那个ListNode和进位相加的值
• 都不为null时，返回 两个ListNode和进位相加的值

public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        return helper(l1,l2,0);
    }

    public ListNode helper(ListNode l1, ListNode l2, int carry){
        if(l1==null && l2==null){
            return carry == 0? null : new ListNode(carry);
        }
        if(l1==null && l2!=null){
            l1 = new ListNode(0);
        }
        if(l2==null && l1!=null){
            l2 = new ListNode(0);
        }
        int sum = l1.val + l2.val + carry;
        ListNode curr = new ListNode(sum % 10);
        curr.next = helper(l1.next, l2.next, sum / 10);
        return curr;
    }
}

迭代法：

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复杂度

时间O(n) 空间(1)

思路

迭代写法相比之下更为晦涩，因为需要处理的分支较多，边界条件的组合比较复杂。过程同样是对齐相加，不足位补0。迭代终止条件是两个ListNode都为null。

注意

• 迭代方法操作链表的时候要记得手动更新链表的指针到next
• 迭代方法操作链表时可以使用一个dummy的头指针简化操作
• 不可以在其中一个链表结束后直接将另一个链表串接至结果中，因为可能产生连锁进位

public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode dummyHead = new ListNode(0);
        if(l1 == null && l2 == null){
            return dummyHead;
        }
        int sum = 0, carry = 0;
        ListNode curr = dummyHead;
        while(l1!=null || l2!=null){
            int num1 = l1 == null? 0 : l1.val;
            int num2 = l2 == null? 0 : l2.val;
            sum = num1 + num2 + carry;
            curr.next = new ListNode(sum % 10);
            curr = curr.next;
            carry = sum / 10;
            l1 = l1 == null? null : l1.next;
            l2 = l2 == null? null : l2.next;
        }
        if(carry!=0){
            curr.next = new ListNode(carry);
        }
        return dummyHead.next;
    }
}

21. Merge Two Sorted Lists

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

思路1:1）考虑特殊情况，两个链表至少有一个为空；

　　   2）考虑两个链表长度不一样的情况；

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {

        if(l1==null)
            return l2;
        if(l2==null)
            return l1;

        ListNode head;
        if(l1.val<=l2.val){
            head = new ListNode(l1.val);
            l1 = l1.next;
        }else{
            head = new ListNode(l2.val);
            l2 = l2.next;
        }
        ListNode node = head;
        while(l1!=null&&l2!=null){
            if(l1.val<=l2.val){
                node.next = l1;
                node = node.next;
                l1 = l1.next;
            }else{
                node.next = l2;
                node = node.next;
                l2 = l2.next;
            }
        }
        if(l1!=null){
            node.next = l1;
        }
        if(l2!=null){
            node.next = l2;
        }
        return head;
    }
}

206. Reverse Linked List

Reverse a singly linked list.

思路：1）首先判断头节点是否为空，为空则直接返回；

　　2）

public class Solution {
    public ListNode reverseList(ListNode head) {
        if(head==null)
            return null;
        ListNode newHead = new ListNode(head.val);
        ListNode p = head.next;
        while(p!=null){
            ListNode node = new ListNode(p.val);
            node.next = newHead;
            p = p.next;
            newHead = node;
        }

        return newHead;
    }
}

sort-list

Sort a linked list in O(n log n) time using constant space complexity.

思路：要求时间复杂度为O(nlogn)，空间复杂度为O(1)；

　　1.考虑使用归并排序；

　　2.归并排序需要找到中点，考虑使用快慢指针；

　　3.归并中排序；

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode sortList(ListNode head) {
        if(head==null||head.next==null)
            return head;
        ListNode mid = getMid(head);
        ListNode right = sortList(mid.next);
        mid.next = null;
        ListNode left = sortList(head);
        return mergeList(left,right);
    }
    public ListNode getMid(ListNode head){
        ListNode slow = head;
        ListNode fast = head.next;
        while(fast!=null&&fast.next!=null){
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }

    public ListNode mergeList(ListNode left,ListNode right){
        if(left==null)
            return right;
        if(right==null)
            return left;
        ListNode head = null;
        if(left.val<=right.val){
            head = left;
            left = left.next;
        }else{
            head = right;
            right = right.next;
        }
        ListNode node = head;
        while(left!=null&&right!=null){
            if(left.val<=right.val){
                node.next = left;
                left = left.next;
            }else{
                node.next = right;
                right = right.next;
            }
            node = node.next;
        }
        if(left!=null){
            node.next = left;
        }
        if(right!=null){
            node.next = right;
        }
        return head;
    }
}

insertion-sort-list

Sort a linked list using insertion sort.

使用插入的方式对链表进行排序：

插入排序的思路如下：当前数与其前面的有序数依次进行比较，找到适当的位置插入，以此类推，得到最终结果；

此题的思路是：

　　1）如果head为空或者head.next为空，则直接返回head;

　　2)使用一个节点cur遍历当前待排序的节点，利用另一个节点pre从头开始遍历，若pre的值<=cur的值且两者不同，pre=pre.next；否则，记录第一个>cur的节点及其值，再遍历此节点到cur节点，调整节点值得位置，将cur的值交换到第一个>cur的节点处，直到链表遍历完；

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode insertionSortList(ListNode head) {
        if(head==null||head.next==null){
            return head;
        }
        ListNode cur = head;
        while(cur!=null){
            ListNode pre = head;
            while(pre.val<=cur.val&&pre!=cur){
                pre = pre.next;
            }
            int firstVal = pre.val;
            ListNode mark = pre;
            while(pre!=cur){
                int nextVal = pre.next.val;
                int tmp = nextVal;
                pre.next.val = firstVal;
                firstVal = tmp;
                pre = pre.next;
            }
            mark.val = firstVal;
            cur = cur.next;
        }
        return head;
    }
}

 

reorder-list

Given a singly linked list L: L0→L1→…→Ln-1→Ln,

reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

You must do this in-place without altering the nodes' values.

For example,
Given{1,2,3,4}, reorder it to{1,4,2,3}.

解题思路：

　　先使用快慢指针找到链表的中点，反转后半部分的链表，再以中点为断点进行前后两部分交叉合并；

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public void reorderList(ListNode head) {
        if(head==null||head.next==null)
            return;
        ListNode fast = head.next;
        ListNode slow = head;
        while(fast!=null&&fast.next!=null){
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode pre = reverseList(slow.next);
        slow.next = pre;
        ListNode p = head;
        ListNode q = slow.next;
        while(q!=null&&p!=null){
            slow.next = q.next;
            q.next = p.next;
            p.next = q;
            p = q.next;
            q = slow.next;
        }
    }
　　//反转单链表
    public ListNode reverseList(ListNode head){
        if(head==null||head.next==null)
            return head;
        ListNode cur = head.next;
        ListNode pre = head;
        while(cur!=null){
            ListNode node = cur.next;
            cur.next = pre;
            pre = cur;
            cur = node;
        }
        head.next = cur;
        return pre;
    }
}