LeetCode 2. Add Two Numbers (暴力)

我就是我 2023-05-31 10:18 47阅读 0赞

题目描述：

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjUyNjE5_size_16_color_FFFFFF_t_70][]

暴力解：

/**
     * 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) {
            //代表是否用过最后一个
            int flagL1 = 0;
            int flagL2 = 0;
            //两个都是1个元素的情况判断
            int twoCase = 0;
            //判断最后一个出循环
            int loopBool = 0;
            //进位数
            int moreAdd = 0;
            //按他的莫名规则来
            // 先把头结点搞了
            ListNode* headNode = new ListNode(0);
            //头结点要保留，新建一个
            ListNode* tmpNode = new ListNode(0);
            tmpNode = headNode;
            //遍历
            while(true){
                int l1EndNum = 0;
                int l2EndNum = 0;
                if(l1->next!=NULL){
                    l1EndNum = l1->val;
                    //后移
                    l1 = l1->next;
                }else{
                    //最后一个
                    if(flagL1 == 0){
                        l1EndNum = l1->val;
                        flagL1 += 1;
                        twoCase += 1;
                    }
                }
                if(l2->next!=NULL){
                    l2EndNum = l2->val;
                    //后移
                    l2 = l2->next;
                }else{
                    //最后一个
                    if(flagL2 == 0){
                        l2EndNum = l2->val;
                        flagL2 += 1;
                        twoCase += 1;
                    }
                }
                int tmpResult = l1EndNum + l2EndNum + moreAdd;
                //进位
                moreAdd = floor(tmpResult / 10);
                int headVal = tmpResult % 10;
                tmpNode->val = headVal;
                //判断[0] [0]情况
                if(twoCase == 2) break;
                if(loopBool == 0){
                    //下一个节点
                    ListNode* nextNode = new ListNode(0);
                    tmpNode->next = nextNode;
                    tmpNode = nextNode;
                }
                if(loopBool == 1) break; 
                if(l1->next==NULL && l2->next==NULL) loopBool = 1;
            }
            
            while(moreAdd != 0){
                //下一个节点
                ListNode* nextNode = new ListNode(0);
                nextNode->val = moreAdd;
                tmpNode->next = nextNode;
                tmpNode = nextNode;
                //置零，要不无限循环
                moreAdd = 0;
            }
            return headNode;
        }
    };

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjUyNjE5_size_16_color_FFFFFF_t_70 1][]

[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjUyNjE5_size_16_color_FFFFFF_t_70]: /images/20230531/d6f2c1feb48e4b249e051f49e574843e.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjUyNjE5_size_16_color_FFFFFF_t_70 1]: /images/20230531/b63a32ba7ef04f7892cc0f32d204a441.png