Leetcode刷题总结: 445. Add Two Numbers II

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Follow up:
What if you cannot modify the input lists? In other words, reversing the lists is not allowed.

Example:

Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 8 -> 0 -> 7

 

思路一：

题目说了代表两个int型的数，第一个思路是我直接将这个list转成int型的正整数， 然后两个int型相加，为了防止溢出，用long存储sum；

再将sum转换成list；

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    unsigned int getNum(ListNode *pnode) {
        unsigned int result = 0;
        while(pnode != NULL) {
            result = result*10 + pnode->val;
            pnode = pnode->next;
        }
        return result;
    }
    
    ListNode* transNumToList(long num) {
        ListNode *cur = new ListNode(num%10);
        num = num/10;
        while (num > 0) {
            int res = num%10;
            num /= 10;
            ListNode *pnew = new ListNode(res);
            pnew->next = cur;
            cur = pnew;
        }
        return cur;
        
    }
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        unsigned int num1 = getNum(l1);
        unsigned int num2 = getNum(l2);
        cout << num1 << endl;
        cout << num2 << endl;
        unsigned long sum = num1+num2;
        return transNumToList(sum);
    }
};

　　在LeetCode上测试，不通过，原因是输入的两个正整数并不是int范围的，数字范围可以是无穷大的；所以不能用直接相加的方式来完成

思路二：

　　只能用链表逐个相加的方式；因为题目说了没有以0开头的除非元素本身就是零，所以

　　1. 分别计算list1和list2的长度为len1和len2,  用vector<int> sum保存和的各个位，则sum的size最大为maxLen+1， pLarge指向l1和l2中len大的listnode, pSmall指向l1和l2中len小的listnode；

　　2. 用一个指针先走lendiff=|len2-len1|步，则sum[0...lendiff] = pLarge[0...lendiff]

　　3. temp = pLarge+pSmall；如果temp>10, 

class Solution {
public:
    int calcListLen(ListNode *pnode) {
        int len = 0;
        while(pnode != NULL) {
            ++len;
            pnode = pnode->next;
        }
        return len;
    }
    
    void addOneToSum(vector<int> &sum, int index) {    
        for (int i = index; i>=0; --i) {
            if (sum[i] == 9) {
                sum[i] = 0;
            }else {
                sum[i] += 1;
                break;
            }
        }
    }
    
    ListNode *transToList(const vector<int> &sum) {
        ListNode *pPrev = new ListNode(-1);
        ListNode *pRoot = pPrev;
        if (sum[0] > 0) {
            pPrev->next = new ListNode(sum[0]);
            pPrev = pPrev->next;
        }
        for (int i = 1; i < sum.size(); ++i) {
            pPrev->next = new ListNode(sum[i]);
            pPrev = pPrev->next;
        }
        return pRoot->next;
    }
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        int len1 = calcListLen(l1);
        int len2 = calcListLen(l2);
        int lendiff = 0;
        ListNode *pLarge = NULL;
        ListNode *pSmall = NULL;
        if (len1 > len2) {
            pLarge = l1;
            pSmall = l2;
            lendiff = len1 -len2;
        }else {
            pLarge = l2;
            pSmall = l1;
            lendiff = len2 - len1;
        }
        
        int maxLen = max(len1, len2);
        vector<int> sum(maxLen+1, 0);
        int sumIndex = 1;  //sum[0]存放进位，当存在的时候
        for (int i = 0; i < lendiff; ++i) {
            sum[sumIndex++] = pLarge->val;
            pLarge = pLarge->next;
        }
        while (pLarge != NULL) {
            int temp = pLarge->val + pSmall->val;
            if (temp >= 10) {
                sum[sumIndex] = temp%10;
                addOneToSum(sum, sumIndex-1);
            }
            else {
                sum[sumIndex] = temp;
            }
            ++sumIndex;
            pLarge = pLarge->next;
            pSmall = pSmall->next;
        }
        return transToList(sum);
    }
};