随心刷题计划
今日题解:
1. Two Sum
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
int n = nums.size();
for (int i = 0; i < n-1;i++){
for (int j = 1; j < n;j++){
if(nums[i]+nums[j]==target)
return {
i, j};
}
}
return {
};
}
};
2. Add Two Numbers
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode *head = nullptr, *tail = nullptr;
int carry = 0;
while (l1 || l2) {
int n1 = l1 ? l1->val: 0;
int n2 = l2 ? l2->val: 0;
int sum = n1 + n2 + carry;
if (!head) {
head = tail = new ListNode(sum % 10);
} else {
tail->next = new ListNode(sum % 10);
tail = tail->next;
}
carry = sum / 10;
if (l1) {
l1 = l1->next;
}
if (l2) {
l2 = l2->next;
}
}
if (carry > 0) {
tail->next = new ListNode(carry);
}
return head;
}
};
- 7. Reverse Integer
class Solution {
public:
int reverse(int x) {
int rev = 0;
while (x != 0) {
if (rev < INT_MIN / 10 || rev > INT_MAX / 10) {
return 0;
}
rev = rev * 10 + x % 10;
x /= 10;
}
return rev;
}
};
- 9. Palindrome Number
class Solution {
public:
bool isPalindrome(int x) {
if(x<0)
return false;
else{
vector<int> a;
while(x){
a.push_back(x % 10);
x /= 10;
}
int flag = 1;
for (int i = 0; i < a.size()/2;i++){
if(a[i]!=a[a.size()-i-1]){
flag = 0;
break;
}
}
if(flag)
return true;
else
return false;
}
}
};
class Solution {
public:
bool isPalindrome(int x) {
if(x<0)
return false;
else{
long a=0,b=x;
while(b){
a=a*10+b%10;
b/=10;
}
return a==x;
}
}
};
- 13. Roman to Integer
class Solution {
public:
int romanToInt(string s) {
unordered_map<char, int> a = {
//无序map
{
'I', 1},
{
'V', 5},
{
'X', 10},
{
'L', 50},
{
'C', 100},
{
'D', 500},
{
'M', 1000},
};
int ans = 0;
int n = s.length();
for (int i = 0; i < n;i++){
int value = a[s[i]];
if (i < n - 1 && value < a[s[i + 1]])
ans -= value;
else
ans += value;
}
return ans;
}
};
- 14. Longest Common Prefix
class Solution {
public:
string longestCommonPrefix(vector<string>& strs) {
if(!strs.size())
return "";
string ans = strs[0];
for (int i = 1; i < strs.size();i++){
string s = ans;
ans = "";
for (int j = 0; j < strs[i].size(); j++){
if(s[j]==strs[i][j])
ans += s[j];
else
break;
}
if(ans=="")
break;
}
return ans;
}
};
- 20. Valid Parentheses
class Solution {
public:
bool isValid(string s) {
stack<char> q;
int n = s.size();
for (int i = 0; i < n;i++){
if(s[i]=='('||s[i]=='{'||s[i]=='['||q.empty())//越界访问注意!!!
q.push(s[i]);
else{
if(s[i]==']'&&q.top()=='[')
q.pop();
else if(s[i]=='}'&&q.top()=='{')
q.pop();
else if(s[i]==')'&&q.top()=='(')
q.pop();
else
q.push(s[i]);
}
}
if(q.empty())
return true;
else
return false;
}
};
- 21. Merge Two Sorted Lists
class Solution{
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
ListNode* head = new ListNode(); //头结点
ListNode* l3 = head; //新链表
while(l1 && l2 ){
if(l1->val >= l2->val){
l3->next = l2;
l2 = l2->next;
}
else{
l3->next = l1;
l1 = l1->next;
}
l3 = l3->next;
}
l3->next = (l1 != NULL) ? l1 : l2;
return head->next;
}
};
class Solution{
public:
ListNode *mergeTwoLists(ListNode *l1, ListNode *l2){
if(l1==nullptr)
return l1;
else if(l2==nullptr)
return l2;
else if(l1->val<l2->val){
l1->next = mergeTwoLists(l1->next, l2);
return l1;
}else{
l2->next = mergeTwoLists(l1, l2->next);
return l2;
}
}
};
- 26. Remove Duplicates from Sorted Array
class Solution {
public:
int removeDuplicates(vector<int>& nums) {
if(nums.size()<2)
return nums.size();
auto it = nums.begin() + 1;
while(it!=nums.end()){
if(*it==*(it-1))
nums.erase(it);
else
it++;
}
return nums.size();
}
};
class Solution {
public:
int removeDuplicates(vector<int>& nums) {
set<int>st(nums.begin(), nums.end());
nums.assign(st.begin(), st.end());
return nums.size();
}
};
class Solution {
public:
int removeDuplicates(vector<int>& nums) {
int size = nums.size();
if (size < 2)
return size;
int slow = 0;//重排指针
for (int fast = 0; fast < size; fast++) {
if (nums[slow] != nums[fast]) {
slow++;
nums[slow] = nums[fast];
}
}
return slow + 1;//[0]位置还有一个元素
}
};
- 27. Remove Element
class Solution {
public:
int removeElement(vector<int>& nums, int val) {
auto it=nums.begin();
while( it != nums.end()){
if(*it==val)
nums.erase(it);
else
it++;
}
return nums.size();
}
};