【LeetCode】0002——兩數相加

題目描述

解題思路

Java代碼

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode res = null; //結果鏈表，不可移動，一旦移動就會丟失值
        ListNode cur = null; //結果鏈表的遊標，可移動，爲了往結果鏈表中添加
        int upval = 0; //進位值
        //情況一
        while(l1 != null && l2 != null){
            int val1 = l1.val;
            int val2 = l2.val;
            int val = val1 + val2 + upval;
            
            upval = val / 10;
            int curval = val % 10;
            
            ListNode thisNode = new ListNode(curval);
            
            if(res == null){
                res = thisNode;
                cur = res;
            }else{
                cur.next = thisNode;
                cur = thisNode;
            }
            l1 = l1.next;
            l2 = l2.next;
        }
        //情況二
        while(l1 == null && l2 != null){  //l1長度小於l2
            int val = l2.val + upval;
            upval = val / 10;
            int curval = val % 10;
            
            ListNode thisNode = new ListNode(curval);
            
            if(res == null){
                res = thisNode;
                cur = res;
            }else{
                cur.next = thisNode;
                cur = thisNode;
            }
            l2 = l2.next;
            
        }
        //情況三
        while(l2 == null && l1 != null){  //l2長度小於l1
            int val = l1.val + upval;
            upval = val / 10;
            int curval = val % 10;
            
            ListNode thisNode = new ListNode(curval);
            
            if(res == null){
                res = thisNode;
                cur = res;
            }else{
                cur.next = thisNode;
                cur = thisNode;
            }
            l1 = l1.next;
        }
        
        if(upval != 0){
            ListNode thisNode = new ListNode(upval);
            cur.next = thisNode;
        }
        
        return res;
    }
}

此代碼還可以整合下，我這麼寫主要是爲了看起來和圖示搭配，可以按下面兩個代碼的方式寫，看起來簡單點

Python3代碼

class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        if l1 is None:
            return l2
        if l2 is None:
            return l1
        
        res = ListNode(0)
        cur = res
        upval = 0
        while l1 or l2:
            tmpsum = 0
            if l1:  # 情況二
                tmpsum = l1.val
                l1 = l1.next
            if l2:  # 情況三
                tmpsum += l2.val
                l2 = l2.next
            curval = (tmpsum + upval) % 10
            upval = (tmpsum + upval) // 10
            cur.next = ListNode(curval)
            cur = cur.next
        # 最後一位是否進位
        if upval:
            cur.next = ListNode(1)
        del cur
        res = res.next
        return res

C++代碼

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        if(l1 == NULL){
            return l2;
        }
        if(l2 == NULL){
            return l1;
        }
        
        ListNode* res = new ListNode(0);
        ListNode* cur = res;
        int upval = 0;
        
        while(l1 != NULL || l2 != NULL){
            int tmpsum = 0;
            if(l1 != NULL){
                tmpsum = l1->val;
                l1 = l1->next;
            }
            if(l2 != NULL){
                tmpsum += l2->val;
                l2 = l2->next;
            }
            int curval = (tmpsum + upval) % 10;
            upval = (tmpsum + upval) / 10;
            cur->next = new ListNode(curval);
            cur = cur->next;
        }
        if(upval != 0){
            cur->next = new ListNode(1);
        }
        res = res->next;
        return res;
    }
};