leetcode problem solve 2——Add Two Numbers

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题意就不再赘述了，这里主要总结一下解题过程中感触比较深的地方。

这道题仅仅是medium水平，但是我却做了两个多小时.....我觉得主要原因是我一开始觉得很简单，然后一上来就写代码，没有做好前期的问题剖析工作，导致后面每次提交都出现不同的案例测试无法通过，然后真的就慌乱了，心想：这种问题我居然都无法控制住？？！！然后心态就崩了....开始焦躁，又试图使自己再次专注解题。做人真的不能骄傲阿，你以为你自己已经足够冷静，足够理性，这只是你以为....真的到了上战场的时候，还是会因为大意丢掉性命。

前期的分析工作一定要做好，特殊情况要抽出来，要和普通情况分离开，特殊的提前讨论，提前解决，解决了之后，再进行一般情况的讨论。

代码：

#include<iostream>#include<vector>using namespace std;struct ListNode{int val;ListNode() { val = 0; next = NULL; }ListNode *next;ListNode(int x) : val(x), next(NULL) {}};class Solution{public:Solution() {}ListNode* addTwoNumbers(ListNode* l1, ListNode* l2){int time = 0;//计算相加次数int tenSignal = 0;ListNode* resultList = new ListNode(1);//用于保存结果ListNode* curNode = resultList;   //当前操作的节点//只有一位数的先讨论if (l1->next == nullptr && l2->next == nullptr){int result = l1->val + l2->val;int x = result - 10;if (x >= 0){resultList->val = x;ListNode* node = new ListNode(1);curNode->next = node;curNode = node;}else{resultList->val = result;}return resultList;}while (l1 != nullptr || l2 != nullptr){int result = 0;if (l1 != nullptr && l2 != nullptr){result = l1->val + l2->val + tenSignal;int x = result - 10;if (x >= 0){tenSignal = 1;result = x;if (time == 0){resultList->val = x;}else{ListNode* node = new ListNode(1);node->val = x;curNode->next = node;curNode = node;}if (l1->next == nullptr && l2->next == nullptr){ListNode* node1 = new ListNode(1);curNode->next = node1;curNode = node1;}}else{tenSignal = 0;if (time == 0){resultList->val = result;}else{ListNode* node = new ListNode(1);node->val = result;curNode->next = node;curNode = node;}}l1 = l1->next;l2 = l2->next;}else if (l1 == nullptr && l2 != nullptr){//l1不空，l2空result = l2->val + tenSignal;int x = result - 10;if (x >= 0){tenSignal = 1;//产生1个节点 ListNode* node2 = new ListNode(0);curNode->next = node2;curNode = node2;if (l2->next == nullptr){ListNode* node3 = new ListNode(1);curNode->next = node3;curNode = node3;}}else{tenSignal = 0;if (time == 0){resultList->val = result;}else{//产生一个节点ListNode* node1 = new ListNode(1);node1->val = result;curNode->next = node1;curNode = node1;}}l2 = l2->next;}else{//l1不空，l2空result = l1->val + tenSignal;int x = result - 10;if ( x >= 0 ){tenSignal = 1;//产生1个节点 ListNode* node2 = new ListNode(0);  curNode->next = node2;curNode = node2;if (l1->next == nullptr){ListNode* node3 = new ListNode(1);curNode->next = node3;curNode = node3;}}else{tenSignal = 0;if (time == 0){resultList->val = result;}else{//产生一个节点ListNode* node1 = new ListNode(1);node1->val = result;curNode->next = node1;curNode = node1;}}l1 = l1->next;}time++;}//end whilecurNode->next = nullptr;return resultList;}//end func};int main(){//构造两条链表ListNode* l1 = new ListNode(3);ListNode* l1_cur = l1;ListNode node2(7);//ListNode node3(9);l1_cur->next = &node2;l1_cur = &node2;/*l1_cur->next = &node3;l1_cur = &node3;*/l1_cur->next = nullptr;ListNode* l2 = new ListNode(9);ListNode* l2_cur = l2;ListNode node5(2);//ListNode node6(3);l2_cur->next = &node5;l2_cur = &node5;/*l2_cur->next = &node6;l2_cur = &node6;*/l2_cur->next = nullptr;//testSolution s;ListNode* result = s.addTwoNumbers(l2, l1);int time = 0;while (result){cout <<"result: " << result->val << endl;result = result->next; //cout << "time = " << ++time<< endl;}}

效果：这次还比较满意

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