week2-leetcode #2-Add two Numbers[Medium]

leetcode #2-Add two Numbers[Medium]

Question

You are given two non-empty linked lists representing two non-negative integers. 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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example

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

Solution1[base]

time complecity: O(n)

space complecity:O(n)

runtime:53ms

class Solution {public:  ListNode* addTwoNumbers(ListNode* l1_head, ListNode* l2_head) {    ListNode* l1_seq = l1_head;    ListNode* l2_seq = l2_head;    ListNode* l3_head = new ListNode(0);    ListNode* l3_seq = l3_head;    int carry = 0;    int sum = 0;    bool flag = true;    while (l1_seq->next != NULL && l2_seq->next != NULL) {      int l1_number = l1_seq->val;      int l2_number = l2_seq->val;      int add = l1_number+l2_number+carry;      sum = add%10;      carry = add/10;      // 第一次      if (flag == true) {        l3_seq->val = sum;        flag = false;      } else {        l3_seq->next = new ListNode(sum);        l3_seq = l3_seq->next;      }      l1_seq = l1_seq->next;      l2_seq = l2_seq->next;    }    // 最后一次两个都可以相加    int l1_number = l1_seq->val;    int l2_number = l2_seq->val;    int last_add = l1_number+l2_number+carry;    sum = last_add%10;    carry = last_add/10;    // 第一次    if (flag == true) {      l3_seq->val = sum;      flag = false;    } else {      l3_seq->next = new ListNode(sum);      l3_seq = l3_seq->next;    }    if (l1_seq->next == NULL && l2_seq->next == NULL && carry != 0) {      l3_seq->next = new ListNode(carry);    }    l1_seq = l1_seq->next;    l2_seq = l2_seq->next;    while (l1_seq != NULL) {      last_add = l1_seq->val+carry;      sum = last_add%10;      carry = last_add/10;      l3_seq->next = new ListNode(sum);      l3_seq = l3_seq->next;      l1_seq = l1_seq->next;      if (l1_seq == NULL && carry != 0) {        l3_seq->next = new ListNode(carry);      }    }    while (l2_seq != NULL) {      last_add = l2_seq->val+carry;      sum = last_add%10;      carry = last_add/10;      l3_seq->next = new ListNode(sum);      l3_seq = l3_seq->next;      l2_seq = l2_seq->next;      if (l2_seq == NULL && carry != 0) {        l3_seq->next = new ListNode(carry);      }    }    return l3_head;  }};

​ 思路：实现考虑了很多种情况：

1. l1有多个，l2有多个，不产生进位

2. l1有多个，l2有多个，产生进位

3. l1有一个，l2有多个，不产生进位

4. l1有一个，l2有多个，产生进位

5. ….

   所以代码很长，逻辑不清楚

Solution2[optimal]

time complecity: O(n)

space complecity:O(n)

runtime:51ms

class Solution {public:  ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {    ListNode preHead(0), *p = &preHead;    int extra = 0;    while (l1 || l2 || extra) {      int sum = (l1 ? l1->val : 0) + (l2 ? l2->val : 0) + extra;      extra = sum / 10;      p->next = new ListNode(sum % 10);      p = p->next;      l1 = l1 ? l1->next : l1;      l2 = l2 ? l2->next : l2;    }  return preHead.next;  }};

思路：将问题简化，考虑l1，l2和进位决定是否增加一个ListNode，虽然复杂度差不多，但是代码量大大减少。逻辑清晰