如何实现数据结构与算法之两数相加

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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 = Noneclass 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);            *t = node;            t = (&node->next);                        carry = sum/10;        }                if (carry > 0) {            ListNode *node = new ListNode(carry);            *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 #include #include /* 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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