数据结构与算法之两数相加

原创
05/16 20:18
阅读数 17

一、说明

         给定两个 非空 链表来表示两个非负整数。位数按照 逆序 方式存储,它们的每个节点只存储单个数字。将两数相加返回一个新的链表。
你可以假设除了数字 0 之外,这两个数字都不会以零开头。
        示例:
                输入: (2 -> 4 -> 3) + (5 -> 6 -> 4)
                输出: 7 -> 0 -> 8
                原因: 342 + 465 = 807

二、解决方案参考

        1. Swift 语言

class AddTwoNumbers {
    func addTwoNumbers(l1: ListNode?, _ l2: ListNode?) -> ListNode? {
        var carry = 0, l1 = l1, l2 = l2
        let dummy = ListNode(0)
        var node = dummy
        
        while l1 != nil || l2 != nil || carry != 0 {
            if l1 != nil {
                carry += l1!.val
                l1 = l1!.next
            }
            if l2 != nil {
                carry += l2!.val
                l2 = l2!.next
            }
            
            node.next = ListNode(carry % 10)
            node = node.next!
            
            carry = carry / 10
        }
        
        return dummy.next 
    }
}

        2. JavaScript 语言


/**
 * 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;
};

        3. Python 语言

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    # maybe standard version
    def _addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        p = dummy = ListNode(-1)
        carry = 0
        while l1 and l2:
            p.next = ListNode(l1.val + l2.val + carry)
            carry = p.next.val / 10
            p.next.val %= 10
            p = p.next
            l1 = l1.next
            l2 = l2.next
        
        res = l1 or l2
        while res:
            p.next = ListNode(res.val + carry)
            carry = p.next.val / 10
            p.next.val %= 10
            p = p.next
            res = res.next
        if carry:
            p.next = ListNode(1)
        return dummy.next
    
    # shorter version
    def addTwoNumbers(self, l1, l2):
        p = dummy = ListNode(-1)
        carry = 0
        while l1 or l2 or carry:
            val = (l1 and l1.val or 0) + (l2 and l2.val or 0) + carry
            carry = val / 10
            p.next = ListNode(val % 10)
            l1 = l1 and l1.next
            l2 = l2 and l2.next
            p = p.next
        return dummy.next
            
        
        

        4. Java 语言

public class ListNode {
    public int val;
    public ListNode next;

    public ListNode(int i) {
        this.val = i;
    }

    public int val() {
        return val;
    }
}
	
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;
}

        5. C++ 语言

class Solution {
    
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
        int x=0, y=0, carry=0, sum=0;
        ListNode *h=NULL, **t=&h;
        
        while (l1!=NULL || l2!=NULL){
            x = getValueAndMoveNext(l1);
            y = getValueAndMoveNext(l2);
            
            sum = carry + x + y;
            
            ListNode *node = new ListNode(sum%10);
            *t = node;
            t = (&node->next);
            
            carry = sum/10;
        }
        
        if (carry > 0) {
            ListNode *node = new ListNode(carry%10);
            *t = node;
        }
        
        return h;
    }
private:
    int getValueAndMoveNext(ListNode* &l){
        int x = 0;
        if (l != NULL){
            x = l->val;
            l = l->next;
        }
        return x;
    }
};

        6. C 语言

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

/* Definition for singly-linked list. */
struct ListNode {
    int val;
    struct ListNode *next;
};

static struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2)
{
    int carry_num = 0;
    int first = 1;
    struct ListNode *res = NULL;
    struct ListNode *p = NULL;
    struct ListNode *prev = p;

    while (l1 != NULL || l2 != NULL || carry_num) {
        int sum = 0;
        int last_carry = carry_num;
        if (l1 != NULL) {
            sum += l1->val;
            l1 = l1->next;
        }
        if (l2 != NULL) {
            sum += l2->val;
            l2 = l2->next;
        }
        if (sum >= 10) {
            sum -= 10;
            carry_num = 1;
        } else {
            carry_num = 0;
        }

        p = malloc(sizeof(*p));
        if (first) {
             res = p;
             first = 0;
        }
        p->val = sum + last_carry;
        if (p->val >= 10) {
            p->val -= 10;
            carry_num = 1;
        }
        p->next = NULL;
        if (prev != NULL) {
             prev->next = p;
        }
        prev = p;
    }

    return res;
}

static struct ListNode *node_build(const char *digits)
{
    struct ListNode *res, *p, *prev;
    int first = 1;
    int len = strlen(digits);
    const char *c = digits + len - 1;
    prev = NULL;
    while (len-- > 0) {
        p = malloc(sizeof(*p));
        if (first) {
            first = 0;
            res = p;
        }
        p->val = *c-- - '0';
        p->next = NULL;
        if (prev != NULL) {
            prev->next = p;
        }
        prev = p;
    }

    return res;
}

static void show(struct ListNode *ln)
{
    int sum = 0, factor = 1;
    while (ln != NULL) {
        sum += ln->val * factor;
        factor *= 10;
        ln = ln->next;
    }
    printf("%d
", sum);
}

int main(int argc, char **argv)
{
    if (argc < 3) {
        fprintf(stderr, "Usage: ./test n1 n2
");
        exit(-1);
    }

    struct ListNode *l1 = node_build(argv[1]);
    struct ListNode *l2 = node_build(argv[2]);
    struct ListNode *res = addTwoNumbers(l1, l2);
    show(l1);
    show(l2);
    show(res);
    return 0;
}

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