408算法题leetcode--第15天

143. 重排链表

143. 重排链表
思路：之前三题的合并
时间：O(n)；空间：O(1)

cpp 复制代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* findMiddle(ListNode* head){
        ListNode* p = head, *q = head;
        // q移动2步，p移动一步
        while(q->next && q->next->next){
            q = q->next->next;
            p = p->next;
        }
        return p;
    }
    ListNode* reverseList(ListNode* head){
        ListNode* pre = nullptr, *cur = head, *next = nullptr;
        while(cur){
            next = cur->next;
            cur->next = pre;
            pre = cur;
            cur = next;
        }
        return pre;
    }
    void mergeList(ListNode* l, ListNode* r){
        ListNode* p = l, *q = r;
        while(p && q){
            ListNode* t1 = p->next, *t2 = q->next;
            p->next = q;
            p = t1;
            q->next = p;
            q = t2;
        }
    }
    void reorderList(ListNode* head) {
        if(head == nullptr) return;
        // 思路：找到链表中点，右链表逆转，逐一合并左右链表节点
        // 1. 找中点
        ListNode* mid_head = findMiddle(head);
        ListNode* right_head = mid_head->next;
        mid_head->next = nullptr;
        // 2. 右边翻转
        right_head = reverseList(right_head);
        // 3. 合并两个链表
        mergeList(head, right_head);
    }
};

2. 两数相加

2. 两数相加
思路：注释
时间：O(max(m, n))；空间：O(1)

cpp 复制代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        // 思路1：读取数字，倒序，相加，倒序，链表
        // 思路2：对应位置相加，进位到后一位
        ListNode* dummy_head = new ListNode();
        ListNode* cur = dummy_head;
        int carry = 0;
        while(l1 || l2 || carry){
            int a = l1 ? l1->val : 0;
            int b = l2 ? l2->val : 0;
            int sum = a + b + carry;
            carry = sum / 10;
            cur->next = new ListNode(sum % 10);
            cur = cur -> next;
            if(l1) l1 = l1->next;
            if(l2) l2 = l2->next;
        }
        return dummy_head->next;
    }
};

445. 两数相加 II

445. 两数相加 II
思路：先翻转，再相加（同上）
时间：O(n)；空间：O(1)

cpp 复制代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head){
        ListNode* pre = nullptr, *cur = head, *next = nullptr;
        while(cur){
            next = cur->next;
            cur->next = pre;
            pre = cur;
            cur = next;
        }
        return pre;
    }
    ListNode* addList(ListNode* lhs, ListNode* rhs){
        ListNode* dummy_head = new ListNode();
        ListNode* cur = dummy_head;
        int carry = 0;
        while(carry || lhs || rhs){
            int a = lhs ? lhs->val : 0;
            int b = rhs ? rhs->val : 0;
            int sum = a + b + carry;
            carry = sum / 10;
            cur->next = new ListNode(sum % 10);
            cur = cur->next;
            if(lhs) lhs = lhs->next;
            if(rhs) rhs = rhs->next;
        }
        return dummy_head->next;
    }
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        // 反转2个链表，然后相加
        ListNode* p = reverseList(l1), *q = reverseList(l2);
        ListNode* ret = addList(p, q);
        return reverseList(ret);
    }
};