Leetcode算法刷题笔记（五）
注：本页有大量的锁定题，由于本人非常穷，所以本页的题非常少。
217. Contains Duplicate
class Solution {
public:
bool containsDuplicate(vector& nums) {
mapa;
for(int i=0;i(nums[i],1));
else return true;
}
return false;
}
}; 注：简单题，想法就是用map做，线性时间复杂度，可是比我快的都是先排序再判断相邻，那就是O(nlog(n))了，真奇了怪了。faster than 17.97%。
218. The Skyline Problem
注：困难题，。faster than XX%。
219. Contains Duplicate II
class Solution {
public:
bool containsNearbyDuplicate(vector& nums, int k) {
mapa;
for(int i=0;i(nums[i],i));
else
{
if(abs(a.find(nums[i])->second-i)<=k)
return true;
else
a.find(nums[i])->second=i;
}
}
return false;
}
};
注：简单题，跟217没啥区别。faster than 18.30%。
220. Contains Duplicate III
class Solution {
public:
bool containsNearbyAlmostDuplicate(vector& nums, int k, int t) {
map m;
int j = 0;
for (int i = 0; i < nums.size(); ++i) {
if (i - j > k) m.erase(nums[j++]);
auto a = m.lower_bound((long long)nums[i] - t);
if (a != m.end() && abs(a->first - nums[i]) <= t) return true;
m[nums[i]] = i;
}
return false;
}
}; https://blog.csdn.net/qq508618087/article/details/50610584
注：中等题，利用map的二叉树结构查找。faster than XX%。
221. Maximal Square
https://blog.csdn.net/xudli/article/details/46371673
注：中等题，。faster than XX%。
222. Count Complete Tree Nodes
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int countNodes(TreeNode* root) {
int k,p=0;
if(!root)
return 0;
queue q;
q.push(root);
while(!q.empty())
{
k=q.size();
while(k--)
{
p++;
if(q.front()->left)
q.push(q.front()->left);
if(q.front()->right)
q.push(q.front()->right);
q.pop();
}
}
return p;
}
};
注：中等题，考察层序遍历。faster than 97.63%。
223. Rectangle Area
class Solution {
public:
int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
long a=max(B,F);
long b=max(A,E);
long c=min(D,H);
long d=min(C,G);
long long square_1=abs(A-C)*abs(B-D);
long long square_2=abs(E-G)*abs(F-H);
int w = max(d-b,long(0));
int h = max(c-a,long(0));
int area=w*h;
return square_1+square_2-area;
}
};注：中等题，这题冷不丁一瞅不会，但仔细一想不就是NMS嘛。。。faster than 16.56%。
224. Basic Calculator
class Solution {
public:
int calculate(string s) {
int res = 0, sign = 1, n = s.size();
stack st;
for (int i = 0; i < n; ++i) {
char c = s[i];
if (c >= '0') {
int num = 0;
while (i < n && s[i] >= '0') {
num = 10 * num + s[i++] - '0';
}
res += sign * num;
--i;
} else if (c == '+') {
sign = 1;
} else if (c == '-') {
sign = -1;
} else if (c == '(') {
st.push(res);
st.push(sign);
res = 0;
sign = 1;
} else if (c == ')') {
res *= st.top(); st.pop();
res += st.top(); st.pop();
}
}
return res;
}
}; 注：困难题，。faster than XX%。
225. Implement Stack using Queues
class MyStack {
public:
/** Initialize your data structure here. */
MyStack() {
}
queue q1;
queue q2;
/** Push element x onto stack. */
void push(int x) {
q1.push(x);
}
/** Removes the element on top of the stack and returns that element. */
int pop() {
while(q1.size()!=1)
{
q2.push(q1.front());
q1.pop();
}
int k = q1.front();
q1.pop();
while(!q2.empty())
{
q1.push(q2.front());
q2.pop();
}
return k;
}
/** Get the top element. */
int top() {
while(q1.size()!=1)
{
q2.push(q1.front());
q1.pop();
}
int k = q1.front();
q2.push(q1.front());
q1.pop();
while(!q2.empty())
{
q1.push(q2.front());
q2.pop();
}
return k;
}
/** Returns whether the stack is empty. */
bool empty() {
return q1.empty();
}
};
/**
* Your MyStack object will be instantiated and called as such:
* MyStack* obj = new MyStack();
* obj->push(x);
* int param_2 = obj->pop();
* int param_3 = obj->top();
* bool param_4 = obj->empty();
*/ 注：简单题，跟232没啥区别，俩来回倒腾。
Runtime: 4 ms, faster than 80.07% of C++ online submissions forImplement Stack using Queues.
Memory Usage: 8.8 MB, less than 72.92% of C++ online submissions forImplement Stack using Queues.
226. Invert Binary Tree
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* invertTree(TreeNode* root) {
if(!root)
return NULL;
TreeNode* clone = new TreeNode(root->val);
clone->left = invertTree(root->right);
clone->right = invertTree(root->left);
return clone;
}
};注：简单题，典型的递归题，没啥好说的。
Runtime: 4 ms, faster than 78.68% of C++ online submissions for Invert Binary Tree.
Memory Usage: 9.6 MB, less than 5.14% of C++ online submissions forInvert Binary Tree.
227. Basic Calculator II
别人的：
class Solution {
public:
int calculate(string s) {
int res = 0, d = 0;
char sign = '+';
stack nums;
for (int i = 0; i < s.size(); ++i) {
if (s[i] >= '0') {
d = d * 10 + s[i] - '0';
}
if ((s[i] < '0' && s[i] != ' ') || i == s.size() - 1) {
if (sign == '+') nums.push(d);
if (sign == '-') nums.push(-d);
if (sign == '*' || sign == '/') {
int tmp = sign == '*' ? nums.top() * d : nums.top() / d;
nums.pop();
nums.push(tmp);
}
sign = s[i];
d = 0;
}
}
while (!nums.empty()) {
res += nums.top();
nums.pop();
}
return res;
}
}; 自己的：
class Solution {
public:
int calculate(string s) {
stack p;
stack c;
int m=0,n;
for(int i=0;i'9'||s[i]<'0')
{
int k=atoi(s.substr(m).c_str());
if(c.empty())
{
p.push(k);
c.push(s[i]);
}
else
{
if((c.top()=='*'||c.top()=='/')||(s[i]=='+'||s[i]=='-'))
{
int q=p.top();
p.pop();
int d=jisuan(q,k,c.top());
c.pop();
while(!c.empty()&&((c.top()=='+'||c.top()=='-')&&(s[i]=='+'||s[i]=='-')))
{
int a=p.top();
p.pop();
d=jisuan(a,d,c.top());
c.pop();
}
p.push(d);
c.push(s[i]);
}
else
{
p.push(k);
c.push(s[i]);
}
}
m=i+1;
}
n=atoi(s.substr(m).c_str());
}
p.push(n);
while(!c.empty())
{
int a=p.top();
p.pop();
int b=p.top();
p.pop();
p.push(jisuan(b,a,c.top()));
c.pop();
}
return p.top();
}
int jisuan(int a,int b,char c)
{
if(c=='+')
return a+b;
if(c=='-')
return a-b;
if(c=='*')
return a*b;
if(c=='/')
return a/b;
}
}; 注：中等题，哔了狗这么慢。。从别人的代码可以看出，如果有负号完全可以把数变为负数省去负号。faster than 0.98%。
228. Summary Ranges
class Solution {
public:
vector summaryRanges(vector& nums) {
if(nums.size()==0)
return {};
vector a;
string s;
for(int i=0;i=nums.size()||nums[i]+1!=nums[i+1])
{
a.push_back(s);
s.clear();
continue;
}
}
if(nums[i]+1!=nums[i+1])
{
s+="->";
s+=to_string(nums[i]);
a.push_back(s);
s.clear();
}
}
return a;
}
}; 注：中等题，没啥说的。faster than 100%。
229. Majority Element II
class Solution {
public:
vector majorityElement(vector& nums) {
vector res;
int m = 0, n = 0, cm = 0, cn = 0;
for (auto &a : nums) {
if (a == m) ++cm;
else if (a ==n) ++cn;
else if (cm == 0) m = a, cm = 1;
else if (cn == 0) n = a, cn = 1;
else --cm, --cn;
}
cm = cn = 0;
for (auto &a : nums) {
if (a == m) ++cm;
else if (a == n) ++cn;
}
if (cm > nums.size() / 3) res.push_back(m);
if (cn > nums.size() / 3) res.push_back(n);
return res;
}
}; http://www.cnblogs.com/lightwindy/p/9736278.html
注：中等题，限制了空间复杂度我就不会了，从题意了解，满足条件的数最多为两个需要用摩尔投票法。faster than XX%。
230. Kth Smallest Element in a BST
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int kthSmallest(TreeNode* root, int k) {
vector a;
stack s;
while(!s.empty()||root)
{
while(root)
{
s.push(root);
root=root->left;
}
root=s.top();
a.push_back(root->val);
if(a.size()==k)
return a.back();
s.pop();
root=root->right;
}
}
}; 注：中等题，考察中序遍历的。faster than 66.34%。
231. Power of Two
class Solution {
public:
bool isPowerOfTwo(int n) {
if(!n)
return false;
while(n)
{
if(n==1)
return true;
if(n%2)
return false;
n/=2;
}
return true;
}
};注：简单题，没啥说的。faster than 99.00%。
232. Implement Queue using Stacks
class MyQueue {
public:
/** Initialize your data structure here. */
MyQueue() {
}
stack s1;
stack s2;
/** Push element x to the back of queue. */
void push(int x) {
s1.push(x);
}
/** Removes the element from in front of queue and returns that element. */
int pop() {
while(!s1.empty())
{
s2.push(s1.top());
s1.pop();
}
int k = s2.top();
s2.pop();
while(!s2.empty())
{
s1.push(s2.top());
s2.pop();
}
return k;
}
/** Get the front element. */
int peek() {
while(!s1.empty())
{
s2.push(s1.top());
s1.pop();
}
int k = s2.top();
while(!s2.empty())
{
s1.push(s2.top());
s2.pop();
}
return k;
}
/** Returns whether the queue is empty. */
bool empty() {
return s1.empty();
}
};
/**
* Your MyQueue object will be instantiated and called as such:
* MyQueue* obj = new MyQueue();
* obj->push(x);
* int param_2 = obj->pop();
* int param_3 = obj->peek();
* bool param_4 = obj->empty();
*/ 注：简单题，用俩stack来回倒腾就行了。
Runtime: 0 ms, faster than 100.00% of C++ online submissions forImplement Queue using Stacks.
Memory Usage: 8.9 MB, less than 32.92% of C++ online submissions forImplement Queue using Stacks.
233. Number of Digit One
class Solution {
public:
int countDigitOne(int n) {
int count=0;
int high=n;
int cur=0,b=1;
while(high>0)
{
cur=high%10;
high/=10;
count+=high*b;
if(cur==1){
count+=n%b+1;
}else if(cur>1){
count+=b;
}
b*=10;
}
return count;
}
};
注：困难题，。faster than 43.93%。
234. Palindrome Linked List
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
bool isPalindrome(ListNode* head) {
if(!head||!head->next)
return true;
ListNode* slow=head;
ListNode* fast=head;
while(fast->next&&fast->next->next)
{
slow=slow->next;
fast=fast->next->next;
}
ListNode* mid=reverse(slow->next);
while(mid)
{
if(mid->val!=head->val)
return false;
mid=mid->next;
head=head->next;
}
return true;
}
ListNode* reverse(ListNode* head){
if(!head->next)
return head;
ListNode* p=NULL;
ListNode* q=head;
ListNode* r=q->next;
while(1)
{
q->next=p;
p=q;
q=r;
r=r->next;
if(!r)
{
q->next=p;
return q;
}
}
}
};注：简单题，我的想法是用快慢指针找到中点，然后把中点后的链表反转，然后头指针和此时的慢指针的下一个节点挨个比较是否相等。faster than 53.03%。
235. Lowest Common Ancestor of a Binary Search Tree
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if(!root)
return NULL;
if(p->valval&&q->valval)
return lowestCommonAncestor(root->left,p,q);
else if(p->val>root->val&&q->val>root->val)
return lowestCommonAncestor(root->right,p,q);
else if(p->val>=root->val&&q->val<=root->val||p->val<=root->val&&q->val>=root->val)
return root;
}
}; 注：简单题，因为是二叉搜索树，所以判断pq和root的大小关系就可以了。faster than 52.95%。
236. Lowest Common Ancestor of a Binary Tree
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if(root==p||root==q||root==NULL)
return root;
TreeNode* left=lowestCommonAncestor(root->left, p, q);
TreeNode* right=lowestCommonAncestor(root->right, p, q);
if(left&&right)
return root;
return left==NULL?right:left;
}
};注：中等题，递归条件弄不明白，这个人的思路是不等于qp的节点都设为NULL。faster than XX%。
237. Delete Node in a Linked List
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
void deleteNode(ListNode* node) {
if (!node || !node->next)
return;
node->val = node->next->val;
node->next = node->next->next;
}
};注：智障题，能不能挑点有意义的题来出？！faster than XX%。
238. Product of Array Except Self
class Solution {
public:
vector productExceptSelf(vector& nums) {
int len = nums.size();
if(0 == len || 1 == len){
vector ret = nums;
return ret;
}
vector ins(len, nums[0]);
vector ret(len, nums[len-1]);
int mul1 = nums[0], mul2 = nums[len-1];
for(int i = 1; i < len; ++i){
mul1 *= nums[i];
mul2 *= nums[len-i-1];
ins[i] = mul1;
ret[len-i-1] = mul2;
}
ret[0] = ret[1];
for(int i = 1; i < len-1; ++i){
ret[i] = ins[i-1] * ret[i + 1];
}
ret[len-1] = ins[len-2];
return ret;
}
};
https://blog.csdn.net/sinat_15723179/article/details/81297382
注：中等题，利用正序和逆序辅助数组来做。faster than XX%。
239. Sliding Window Maximum
class Solution {
public:
vector maxSlidingWindow(vector& nums, int k) {
vector a;
int maximum=INT_MIN;
if(nums.size()==0)
return a;
for(int i=0;i<=nums.size()-k;i++)
{
if(i!=0&&nums[i+k-1]>=maximum)
maximum=nums[i+k-1];
else if(i!=0&&nums[i-1]!=maximum);
else
{
maximum=INT_MIN;
for(int j=i;j 注：困难题，本应该是困难题里少有的简单题了，滑窗法，判断最大值的位置即可。faster than 43.93%。
240. Search a 2D Matrix II
别人的：
class Solution {
public:
bool searchMatrix(vector>& matrix, int target) {
if(matrix.empty()||matrix[0].empty()) return false;
int m=matrix.size(),n=matrix[0].size();
int r=0,c=n-1;
while(r=0){
if(matrix[r][c]==target) return true;
else if(matrix[r][c] 自己的代码：
class Solution {
public:
bool searchMatrix(vector>& matrix, int target) {
if(matrix.size()==0||matrix[0].size()==0)
return false;
int low=matrix[0].size(),high,mid,hang,lie;
int p=1,q=2;
while(p!=q)
{
lie=low-1;
low=0,high=matrix.size()-1;
while(low<=high)
{
mid=(high+low)/2;
if(matrix[mid][lie]target)
high=mid-1;
else return true;
}
if(low>matrix.size()-1)
break;
q=matrix[low][lie];
hang=low;
low=0,high=matrix[0].size()-1;
while(low<=high)
{
mid=(high+low)/2;
if(matrix[hang][mid]>target)
high=mid-1;
else if(matrix[hang][mid]matrix[0].size()-1||low==0)
break;
p=matrix[hang][low-1];
}
return false;
}
}; 注：中等题，我的想法是从最后一列开始，二分法判断，找到了就找到了，没找到的判断条件是会停留在同一个数上不变。faster than 24.75%。
241. Different Ways to Add Parentheses
class Solution {
public:
vector diffWaysToCompute(string input) {
vector res;
//边界条件是如果找不到运算符，说明只有
for(int i=0;i 注：中等题，不会不会。faster than XX%。
242. Valid Anagram
class Solution {
public:
bool isAnagram(string s, string t) {
sort(s.begin(),s.end());
sort(t.begin(),t.end());
return s==t;
}
};注：简单题，可以遍历可以排序，排序的肯定慢，但我当然挑字少的写了。faster than 14.51%。
257. Binary Tree Paths
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector binaryTreePaths(TreeNode* root) {
vector a;
string s;
if(!root)
return a;
digui(root,a,s);
return a;
}
void digui(TreeNode* root,vector& a,string s){
if((!root->left)&&(!root->right))
{
s+=to_string(root->val);
a.push_back(s);
return ;
}
s+=to_string(root->val);
if(root->left)
digui(root->left,a,s+"->");
if(root->right)
digui(root->right,a,s+"->");
}
}; 注：简单题，相当于按子树遍历。faster than 33.10%。
258. Add Digits
class Solution {
public:
int addDigits(int num) {
while(num/10)
{
int a=0;
while(num)
{
a=a+num%10;
num/=10;
}
num=a;
}
return num;
}
};注：简单题，没啥说的。faster than 99.29%。
260. Single Number III
class Solution {
public:
vector singleNumber(vector& nums) {
map m;
vector a;
for (int i = 0; i < nums.size(); i++) {
// key 为数组中的值 value为数组中的值出现的次数
if (!m.count(nums[i])) {
m[nums[i]] = 1;
}else {
m[nums[i]] += 1;
}
}
for (int i = 0; i < nums.size(); i++) {
if (m[nums[i]] == 1) {
a.push_back(nums[i]);
}
}
return a;
}
}; 注：中等题，只会哈希表法，位运算是真的弄不明白。faster than 2.64%。
263. Ugly Number
class Solution {
public:
bool isUgly(int num) {
if(!num)
return false;
while(num!=1)
{
if(!(num%2))
num/=2;
else if(!(num%3))
num/=3;
else if(!(num%5))
num/=5;
else return false;
}
return true;
}
};注：简单题，按题意正常做就好了。faster than 97.91%。
264. Ugly Number II
超时代码：
class Solution {
public:
int nthUglyNumber(int n) {
map a;
a[1]=1;
int sum=1,k=2;
while(sum!=n)
{
int num=k;
while(num!=1)
{
if(!(num%2))
num/=2;
else if(!(num%3))
num/=3;
else if(!(num%5))
num/=5;
else
break;
if(a.find(num)!=a.end())
{
a[k]=1;
sum++;
break;
}
break;
}
k++;
}
return --k;
}
}; 别人AC代码：
class Solution {
public:
int nthUglyNumber(int n) {
static int arr[1691] = {[0]= 1},
i = 1, i2 = 0, i3 = 0, i5 = 0,
n2 = 2, n3 = 3, n5 = 5;
for (; i < n; i++) {
arr[i] = min(min(n2, n3), n5);
if (arr[i] == n2) n2 = 2 * arr[++i2];
if (arr[i] == n3) n3 = 3 * arr[++i3];
if (arr[i] == n5) n5 = 5 * arr[++i5];
}
return arr[n-1];
}
};注：中等题，如果不限制内存的话完全可以弄，但是限制了之后这代码还限时，我死也编不出来了。。。faster than XX%。
Rrui的Leetcode算法刷题笔记（六）链接如下：
https://blog.csdn.net/Rrui7739/article/details/83785403