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  • Algorithm——Add Two Numbers

    一、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
    Explanation: 342 + 465 = 807.

    二、solution

    (cpp)

    /**
     * Definition for singly-linked list.
     * struct ListNode {
     *     int val;
     *     ListNode *next;
     *     ListNode(int x) : val(x), next(NULL) {}
     * };
     */
    class Solution {
    public:
        ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode preHead(0), *p = &preHead;
        int carry=0;
            while(l1||l2||carry){
                int sum = (l1?l1->val:0)+(l2?l2->val:0)+carry;
                carry = sum/10;
                p->next = new ListNode(sum % 10);
                p = p->next;
                l1 = l1?l1->next:0;
                l2 = l2?l2->next:0;
            }
            return preHead.next;    
        }
    };

    (js)

    /**
     * Definition for singly-linked list.
     * function ListNode(val) {
     *     this.val = val;
     *     this.next = null;
     * }
     */
    /**
     * @param {ListNode} l1
     * @param {ListNode} l2
     * @return {ListNode}
     */
    
    var addTwoNumbers = function(l1, l2) {
      var add = 0
        , ans
        , head;
    
      while(l1 || l2) {
        var a = l1 ? l1.val : 0
          , b = l2 ? l2.val : 0;
    
        var sum = a + b + add;
        add = ~~(sum / 10);
    
        var node = new ListNode(sum % 10);
    
        if (!ans)
          ans = head = node;
        else {
          head.next = node;
          head = node; 
        }
        
        if (l1)
          l1 = l1.next;
        if (l2)
          l2 = l2.next;
      }
    
      if (add) {
        var node = new ListNode(add);
        head.next = node;
        head = node;
      }
    
      return ans;
    };
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  • 原文地址:https://www.cnblogs.com/bbcfive/p/10748295.html
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