LeetCode笔记

Leetcode笔记

Two Sum
- C（初始版）
- C++版1
- C++版2
- 笔记
Add Two Numbers
- 初始版本
- C

Two Sum

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

C（初始版）

#include <iostream>
int main(int argc, const char * argv[]) {int nums[4]={2,7,11,15};int target;scanf("%d",&target);for(int i=0;i<4;i++){for(int j=i+1;j<4;j++){if(nums[i]+nums[j]==target){printf("%d %d",i,j);return 0;}}}return 0;
}

C++版1

链接： [link]https://www.cnblogs.com/grandyang/p/4130379.html

class Solution {
public:vector<int> twoSum(vector<int>& nums, int target) {unordered_map<int, int> m;vector<int> res;for (int i = 0; i < nums.size(); ++i) {m[nums[i]] = i;}for (int i = 0; i < nums.size(); ++i) {int t = target - nums[i];if (m.count(t) && m[t] != i) {res.push_back(i);res.push_back(m[t]);break;}}return res;}
};

C++版2

class Solution {
public:vector<int> twoSum(vector<int>& nums, int target) {unordered_map<int, int> m;for (int i = 0; i < nums.size(); ++i) {if (m.count(target - nums[i])) {return {i, m[target - nums[i]]};}m[nums[i]] = i;}return {};}
};

笔记

vector count：返回元素值为target的元素个数。

Add Two Numbers

You are given two non-empty linked lists representing two non-negative integers. 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.

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

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807

初始版本

将两个链表的对应项相加，再从头节点遍历新链表（若该节点val>9，则-10，且next的val+1）
代码暂未修改

 /*** 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) {ListNode now1=l1,now2=l2;ListNode *newlist,pnode;pnode=newlist;while(now1!=NULL && now2!=NULL){pnode.val=now1.val+now2.val;now1=now1->next;now2=now2->next;pnode=pnode->next;}if(now1==NULL){pnode->next=now2;}else{pnode->next=now1;}pnode=newlist;while(pnode!=NULL){if(pnode>9){pnode.val-=10;if(pnode->next!=NULL){pnode->next.val+=1;}else if(pnode->next==NULL){pnode->next.val=1;}}}return newlist;}};

C

链接：https://www.cnblogs.com/JeroZeng/p/4668784.html

比较过程中利用carry实时进位，相较于初始版本中再次遍历链表，节省了时间。

struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {struct ListNode head, *p;  p = &head;int carry = 0;while(1){if(l1 && l2){l1->val += (l2->val + carry);carry = l1->val / 10;l1->val = l1->val % 10;p->next = l1;p = l1;l1 = l1->next;l2 = l2->next;}else if(l1){l1->val += carry;carry = l1->val / 10;l1->val = l1->val % 10;p->next = l1;p = l1;l1 = l1->next;}else if(l2){l2->val += carry;carry = l2->val / 10;l2->val = l2->val % 10;p->next = l2;p = l2;l2 = l2->next;}else if(carry){struct ListNode *cur = (struct ListNode*)malloc(sizeof(struct ListNode));cur->val = 1;cur->next = NULL;p->next = cur;return head.next;}elsereturn head.next;}return NULL;
}

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