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linear-list/linked-list/add-two-numbers

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2023-12-01

Add Two Numbers

描述

You are given two linked lists representing two non-negative numbers. 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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)

Output: 7 -> 0 -> 8

分析

Add Binary 很类似

代码

// Add Two Numbers
// 跟Add Binary 很类似
// 时间复杂度O(m+n),空间复杂度O(1)
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(-1); // 头节点
        int carry = 0;
        ListNode prev = dummy;
        for (ListNode pa = l1, pb = l2;
             pa != null || pb != null;
             pa = pa == null ? null : pa.next,
             pb = pb == null ? null : pb.next,
             prev = prev.next) {
            final int ai = pa == null ? 0 : pa.val;
            final int bi = pb == null ? 0 : pb.val;
            final int value = (ai + bi + carry) % 10;
            carry = (ai + bi + carry) / 10;
            prev.next = new ListNode(value); // 尾插法
        }
        if (carry > 0)
            prev.next = new ListNode(carry);
        return dummy.next;
    }
};
// Add Two Numbers
// 跟Add Binary 很类似
// 时间复杂度O(m+n),空间复杂度O(1)
class Solution {
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
        ListNode dummy(-1); // 头节点
        int carry = 0;
        ListNode *prev = &dummy;
        for (ListNode *pa = l1, *pb = l2;
             pa != nullptr || pb != nullptr;
             pa = pa == nullptr ? nullptr : pa->next,
             pb = pb == nullptr ? nullptr : pb->next,
             prev = prev->next) {
            const int ai = pa == nullptr ? 0 : pa->val;
            const int bi = pb == nullptr ? 0 : pb->val;
            const int value = (ai + bi + carry) % 10;
            carry = (ai + bi + carry) / 10;
            prev->next = new ListNode(value); // 尾插法
        }
        if (carry > 0)
            prev->next = new ListNode(carry);
        return dummy.next;
    }
};

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