目錄
1. 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].
思想：簡單查找的想法，還可以用<map
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
for(int i=0;i<nums.size();i++){
for(int j=i+1;j<nums.size();j++){
if(target-nums[i]==nums[j]){
return {i,j};
}
}
}
}
};
2. 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.
/**
* 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) {
if(l1==nullptr)return l1;
if(l2==nullptr)return l2;
ListNode *r=new ListNode(-1),*res=r;//*res=r等價於 ListNode *res;res=r;
ListNode *q=l1,*p=l2;
int carry=0;
while(p||q||carry){
int sum=carry;
if(p){
sum+=p->val;
p=p->next;
}
if(q){
sum+=q->val;
q=q->next;
}
r->next=new ListNode(sum%10);
carry=sum/10;
r=r->next;
}
return res->next;
}
};
3. Longest Substring Without Repeating Characters
Given a string, find the length of the longest substring without repeating characters.
Example 1:
Input: “abcabcbb”
Output: 3
Explanation: The answer is “abc”, with the length of 3.
Example 2:
Input: “bbbbb”
Output: 1
Explanation: The answer is “b”, with the length of 1.
Example 3:
Input: “pwwkew”
Output: 3
Explanation: The answer is “wke”, with the length of 3.
Note that the answer must be a substring, “pwke” is a subsequence and not a substring.
思想：使用set容器，設置左右邊界，出現重複元素，擦除最左元素，並更新max值，未出現不斷加入倒容器中。
class Solution {
public:
int lengthOfLongestSubstring(string s) {
int len=s.length(),max=0;
set<char> sp;
int left=0,right=0;
while(right<len){
if(sp.find(s[right])==sp.end()){
sp.insert(s[right]);
right++;
}else{
if(sp.size()>max){
max=sp.size();
}
sp.erase(s[left]);
left++;
}
}
return max>sp.size()?max:sp.size();
}
};
