151. Reverse Words in a String
class Solution {
public:
void reverseWords(string &s) {
if(s.empty())
return ;
string k,p;
for(int i=0;i<s.size();i++)
{
if(s[i]==' ')
{
if(!p.empty())
{
k=" "+p+k;
p.clear();
}
continue;
}
p+=s[i];
}
if(!p.empty())
k=p+k;
else if(!k.empty())
k=k.substr(1);
s=k;
}
};注：中等题，这题有病，还不让测试的。faster than 31.57%。
152. Maximum Product Subarray
class Solution {
public:
int maxProduct(vector<int>& nums) {
int n=nums.size();
vector<int> maxnum(n,0),minnum(n,0);
int maxres=INT_MIN;
for(int i=0;i<n;i++){
if(i==0){
maxnum[i]=nums[i];
minnum[i]=nums[i];
}
else{
maxnum[i]=max(maxnum[i-1]*nums[i],max(minnum[i-1]*nums[i],nums[i]));
minnum[i]=min(maxnum[i-1]*nums[i],min(minnum[i-1]*nums[i],nums[i]));
}
maxres=max(maxres,maxnum[i]);
}
return maxres;
}
};
https://www.cnblogs.com/ganganloveu/p/4089925.html
https://www.cnblogs.com/bakari/p/4007368.html
注：中等题，典型的动态规划，不会。faster than XX%。
153. Find Minimum in Rotated Sorted Array
class Solution {
public:
int findMin(vector<int>& nums) {
int low=0,high=nums.size()-1;
while(high>=low)
{
int mid=(high+low)/2;
if(nums[mid]>nums[low])
{
if(nums[low]>nums[high])
low=mid+1;
else high=mid-1;
}
else
{
if(mid-1>=low&&nums[mid]<nums[mid-1])
return nums[mid];
else if(mid-1>=low&&nums[mid]>nums[mid-1])
high=mid-1;
else if(mid+1<=high&&nums[mid]>nums[mid+1])
low=mid+1;
else return nums[mid];
}
}
return nums[low];
}
};注：中等题，与第33题类似，主要是找有序区间。faster than 48.15%。
154. Find Minimum in Rotated Sorted Array II
class Solution {
public:
int findMin(vector<int>& nums) {
int low=0,high=nums.size()-1;
while(high>=low)
{
int mid=(high+low)/2;
if(nums[mid]==nums[low]&&mid!=low)
low++;
else if(nums[mid]>nums[low])
{
if(nums[low]>=nums[high])
low=mid+1;
else high=mid-1;
}
else
{
if(mid-1>=low&&nums[mid]<nums[mid-1])
return nums[mid];
else if(mid-1>=low&&nums[mid]>=nums[mid-1])
high=mid-1;
else if(mid+1<=high&&nums[mid]>=nums[mid+1])
low=mid+1;
else return nums[mid];
}
}
return nums[low];
}
};注：困难题，上一题的加强版，加个判断条件，以前的判断条件都换成等号就行了。faster than 99.05%。
155. Min Stack
class MinStack {
public:
/** initialize your data structure here. */
vector<int> a;
int min=INT_MAX;
MinStack(){}
void push(int x) {
a.push_back(x);
}
void pop() {
a.pop_back();
}
int top() {
return a[a.size()-1];
}
int getMin() {
int k=min;
for(int i=0;i<a.size();i++)
if(a[i]<k)
k=a[i];
return k;
}
};
/**
* Your MinStack object will be instantiated and called as such:
* MinStack obj = new MinStack();
* obj.push(x);
* obj.pop();
* int param_3 = obj.top();
* int param_4 = obj.getMin();
*/注：简单题，懒得说了。faster than 5.29%。
160. Intersection of Two Linked Lists
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
ListNode *p=headA,*q=headB;
int n=1,m=1;
if(!p||!q)
return NULL;
while(p->next)
{
p=p->next;
n++;
}
while(q->next)
{
q=q->next;
m++;
}
if(q!=p)
return NULL;
while(--n&&--m)
{
ListNode *c=headA,*d=headB;
int a=n-1,b=m-1;
while(a--)
c=c->next;
while(b--)
d=d->next;
if(c!=d)
return c->next;
}
if(!n)
return headA;
else return headB;
}
};注：简单题，纯暴力解法，优解与前面第141题判断是否有环很相似。faster than 46.84%。
162. Find Peak Element
class Solution {
public:
int findPeakElement(vector<int>& nums) {
int i = 0;
int j = nums.size() - 1;
while(i < j)
{
int temp = (i + j) / 2;
if(nums[temp] > nums[temp + 1])
{
j = temp;
}
else
{
i = temp + 1;
}
}
return i;
}
};
http://courses.csail.mit.edu/6.006/spring11/lectures/lec02.pdf
注：中等题，纯暴力解法，其实要求的复杂度一看就是2分法，2分法思路链接如上，非常聪明。faster than 16.51%。
164. Maximum Gap
class Solution {
public:
int maximumGap(vector<int>& nums) {
if(nums.size()<=1)
return 0;
sort(nums.begin(),nums.end());
int dif=0,max=INT_MIN;
for(int i=1;i<nums.size();i++)
{
dif=nums[i]-nums[i-1];
if(dif>max)
max=dif;
}
return max;
}
};注：困难题，sort一下就行了，线性时间的不会。faster than 99.05%。
165. Compare Version Numbers
class Solution {
public:
int compareVersion(string version1, string version2) {
int i=0,j=0;
while(i<version1.size()||j<version2.size())
{
int a=0,b=0;
while(i<version1.size()&&version1[i]!='.')
a=10*a+version1[i++]-'0';
while(j<version2.size()&&version2[j]!='.')
b=10*b+version2[j++]-'0';
if(a<b)
return -1;
else if(a>b)
return 1;
i++;
j++;
}
return 0;
}
};注：中等题，没啥难度。faster than 100%。
166. Fraction to Recurring Decimal
class Solution {
public:
string fractionToDecimal(int numerator, int denominator) {
long long t = numerator, d = denominator;
map<long long, int> A;
string ans;
if (t*d < 0) ans = "-";
t = abs(t), d = abs(d);
ans += to_string(t / d);
t %= d;
if (!t) return ans;
ans += '.';
while (t) {
if (A.count(t)) {
ans.insert(A[t], "("), ans.push_back(')');
return ans;
}
A[t] = ans.size(), ans += '0' + t * 10 / d;
t = t * 10 % d;
}
return ans;
}
};
注：中等题，考察循环小数判定，出现重复的数字就是循环小数了，利用哈希表。faster than XX%。
167. Two Sum II - Input array is sorted
class Solution {
public:
vector<int> twoSum(vector<int>& numbers, int target) {
int left=0,right=numbers.size()-1;
while(right>left)
{
if(numbers[left]+numbers[right]>target)
right--;
else if(numbers[left]+numbers[right]<target)
left++;
else return {left+1,right+1};
}
return {};
}
};注：简单题，既然已经是升序排列的数组，完全可以双指针法求解。faster than 99.06%。
168. Excel Sheet Column Title
class Solution {
public:
string convertToTitle(int n) {
string s;
if((n-1)/26>=1)
{
int k=(n-1)/26;
s+=convertToTitle(k);
}
s=s+char('A'+(n-1)%26);
return s;
}
};注：简单题，递归求解就好了，唯一需要注意的问题就是字母是从1开始计算的。faster than 100%。
169. Majority Element
class Solution {
public:
int majorityElement(vector<int>& nums) {
map<int,int> a;
for(int i=0;i<nums.size();i++)
{
if(a.find(nums[i])==a.end())
a[nums[i]]=1;
else ++a[nums[i]];
}
int min=0,p;
for(map<int, int>::iterator i=a.begin();i!=a.end();i++)
{
if(i->second>min)
{
min=i->second;
p=i->first;
}
}
return p;
}
};注：简单题，用的最直接的map计数方法，最优解是一次遍历计数。faster than 13.41%。
171. Excel Sheet Column Number
class Solution {
public:
int titleToNumber(string s) {
int sum=0;
for(int i=0;i<s.size();i++)
sum=sum*26+s[i]-'A'+1;
return sum;
}
};注：简单题，没啥好说的。faster than 98.41%。
172. Factorial Trailing Zeroes
class Solution {
public:
int trailingZeroes(int n) {
int count = 0;
while (n > 1)
count += (n /= 5);
return count;
}
};
注：简单题，找规律。faster thanXX%。
173. Binary Search Tree Iterator
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class BSTIterator {
public:
vector<int> a;
int p=0;
BSTIterator(TreeNode* root) {
if(root)
{
stack<TreeNode*> s;
TreeNode* p=root;
while(p||!s.empty())
{
while(p)
{
s.push(p);
p=p->left;
}
p=s.top();
s.pop();
a.push_back(p->val);
p=p->right;
}
}
}
/** @return the next smallest number */
int next() {
return a[p++];
}
/** @return whether we have a next smallest number */
bool hasNext() {
return p<a.size();
}
};
/**
* Your BSTIterator object will be instantiated and called as such:
* BSTIterator* obj = new BSTIterator(root);
* int param_1 = obj->next();
* bool param_2 = obj->hasNext();
*/注：中等题，其实就是二叉树中序遍历。faster than 1.2%。
174. Dungeon Game
注：困难题，。faster than XX%。
179. Largest Number
自己debug版：
class Solution {
private:
static bool key(int a,int b)
{
string p=to_string(a);
string q=to_string(b);
int i=0,j=0;
while(i<p.size()||j<q.size())
{
if(i>=p.size())
{
while(j<q.size()&&(q[0]==q[j]))
j++;
if(j>=q.size()||q[0]==q[j])
return true;
return q[0]>q[j];
}
else if(j>=q.size())
return p[0]<p[i];
else if(p[i]!=q[j])
return p[i]>q[j];
i++;
j++;
}
return false;
}
public:
string largestNumber(vector<int>& nums) {
sort(nums.begin(),nums.end(),key);
string s;
for(int i=0;i<nums.size();i++)
s+=to_string(nums[i]);
if(s[0]=='0')
{
s.clear();
s+='0';
}
return s;
}
};别人AC版：
class Solution {
public:
string largestNumber(vector<int>& nums) {
vector<string> numStr;
for(int num: nums)
numStr.push_back(to_string(num));
sort(numStr.begin(), numStr.end(), cmp);
if(numStr.size() && numStr[0] == "0") return "0";
string s = "";
for(string num: numStr)
s += num;
return s;
}
static bool cmp(string i, string j){
return i + j > j + i;
}
};注：中等题，重点是判断两数谁打谁小那，我写的太麻烦了，自己都绕进去了。faster than XX%。
187. Repeated DNA Sequences
class Solution {
public:
vector<string> findRepeatedDnaSequences(string s) {
unordered_map<string,int> a;
vector<string> b;
if(s.size()<10)
return b;
for(int i=0;i<=s.size()-10;i++)
{
string p=s.substr(i,10);
if(a.find(p)==a.end())
a[p]=1;
else if(a[p]==1)
{
b.push_back(p);
a[p]++;
}
}
return b;
}
};注：中等题，哈希表的应用。faster than 24.31%。
188. Best Time to Buy and Sell Stock IV
注：困难题，。faster than XX%。
189. Rotate Array
class Solution {
public:
void rotate(vector<int>& nums, int k) {
int size=nums.size();
int n=k%size;
for(int i=0;i<size-n;i++)
{
int a=nums[0];
nums.erase(nums.begin());
nums.push_back(a);
}
}
};注：简单题，没意思。faster than 15.95%。
190. Reverse Bits
class Solution {
public:
uint32_t reverseBits(uint32_t n) {
uint32_t k;
int a=32;
while(a--)
{
k=k*2+n%2;
n/=2;
}
return k;
}
};注：简单题，2进制操作。faster than 87.18%。
191. Number of 1 Bits
class Solution {
public:
int hammingWeight(uint32_t n) {
int k=0;
while(n!=0)
{
if(n%2)
k++;
n/=2;
}
return k;
}
};注：简单题，二进制找1。faster than 75.60%。
198. House Robber
记忆化搜索：
class Solution {
public:
int rob(vector<int>& nums) {
vector<int> a(nums.size(),-1);
return digui(nums,a,0);
}
int digui(vector<int>& nums,vector<int>& a,int k)
{
if(k>=nums.size())
return 0;
if(a[k]!=-1)
return a[k];
a[k]=max(digui(nums,a,k+2)+nums[k],digui(nums,a,k+1));
return a[k];
}
};动态规划：
class Solution {
public:
int rob(vector<int>& nums) {
vector<int> a(nums.size()+1,0);
for(int i=nums.size()-1;i>=0;i--)
{
if(i==nums.size()-1)
a[i]=nums[i];
else
a[i]=max(a[i+1],nums[i]+a[i+2]);
}
return a[0];
}
};注：简单题，典型的动态规划。faster than 35.93%。
199. Binary Tree Right Side View
/**
* 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<int> rightSideView(TreeNode* root) {
vector<int> a;
queue<TreeNode*> p;
if(!root)
return a;
p.push(root);
while(!p.empty())
{
int size=p.size();
TreeNode* q;
while(size--)
{
q=p.front();
if(q->left)
p.push(q->left);
if(q->right)
p.push(q->right);
p.pop();
}
a.push_back(q->val);
}
return a;
}
};注：中等题，其实就是二叉树层序遍历只记录每一层的最后一个。faster than 47.86%。
200. Number of Islands
class Solution {
public:
int numIslands(vector<vector<char>>& grid) {
if(grid.size()==0||grid[0].size()==0)
return 0;
vector<vector<int>> a(grid.size(),vector<int>(grid[0].size(),0));
int p=0;
for(int i=0;i<grid.size();i++)
for(int j=0;j<grid[0].size();j++)
if(grid[i][j]=='1')
{
if(a[i][j]==1)
continue;
else
{
digui(a,grid,i,j);
p++;
}
}
return p;
}
void digui(vector<vector<int>>& a,vector<vector<char>>& grid,int i,int j)
{
a[i][j]=1;
if(i-1>=0&&grid[i-1][j]=='1'&&a[i-1][j]==0)
digui(a,grid,i-1,j);
if(j-1>=0&&grid[i][j-1]=='1'&&a[i][j-1]==0)
digui(a,grid,i,j-1);
if(j+1<grid[0].size()&&grid[i][j+1]=='1'&&a[i][j+1]==0)
digui(a,grid,i,j+1);
if(i+1<grid.size()&&grid[i+1][j]=='1'&&a[i+1][j]==0)
digui(a,grid,i+1,j);
}
};注：中等题，我用的深度优先搜索，与79题想法相同，代码就是抄过来修改的。faster than 55.82%
201. Bitwise AND of Numbers Range
class Solution {
public:
int rangeBitwiseAnd(int m, int n) {
int i = 0;
while (m != n) {
m >>= 1;
n >>= 1;
++i;
}
return (m << i);
}
};注：中等题，不会。faster than XX%。
202. Happy Number
class Solution {
public:
bool isHappy(int n) {
int p=0;
while(1)
{
int a=n,k=0;
while(a)
{
k+=pow(a%10,2);
a/=10;
}
if(k==1)
return true;
n=k;
p++;
if(p>10000)
return false;
}
}
};注：简单题，他们找规律说如果不得1的话会得到循环，我没找到这种规律，直接就用的风骚法。。faster than 0.99%。
203. Remove Linked List Elements
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* removeElements(ListNode* head, int val) {
ListNode *q=new ListNode(0);
ListNode* p=q;
p->next=head;
while(p->next)
{
if(p->next->val==val)
{
p->next=p->next->next;
continue;
}
p=p->next;
}
return q->next;
}
};注：简单题，没啥说的。faster than 31.52%。
204. Count Primes
class Solution {
public:
int countPrimes(int n) {
int sum=0;
vector<int> a(n,0);
for(int i=2;i<n;i++)
{
if(!a[i])
{
sum++;
for(int j=2;j*i<n;j++)
a[j*i]=1;
}
}
return sum;
}
};
注：简单题，这题用哈希表要超时，只能用vector，j从2开始取值。faster than 54.51%。
205. Isomorphic Strings
class Solution {
public:
bool isIsomorphic(string s, string t) {
if(s.size()!=t.size())
return false;
unordered_map<char,char> a,b;
for(int i=0;i<s.size();i++)
{
if(a.find(s[i])==a.end())
{
a[s[i]]=t[i];
}
else if(a[s[i]]!=t[i])
return false;
if(b.find(t[i])==b.end())
{
b[t[i]]=s[i];
}
else if(b[t[i]]!=s[i])
return false;
}
return true;
}
};注：简单题，用两个map杜绝一个字母对应两个索引的情况。faster than 30.70%。
206. Reverse Linked List
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* reverseList(ListNode* head) {
ListNode* p=NULL;
ListNode* q=head;
if(!head)
return NULL;
ListNode* r=head->next;
while(r)
{
q->next=p;
p=q;
q=r;
r=r->next;
if(!r)
q->next=p;
}
return q;
}
};注：简单题，经典的反转链表，不会编的话在纸上画个链表就会了。faster than 47.99%。
207. Course Schedule
注：中等题，本题其实有个无比风骚的代码。正常的是用map做的。faster than 37.46%。
208. Implement Trie (Prefix Tree)
注：中等题，本题其实有个无比风骚的代码。正常的是用map做的。faster than 37.46%。
209. Minimum Size Subarray Sum
class Solution {
public:
int minSubArrayLen(int s, vector<int>& nums) {
int sum=0,left=0,right=0,min1=INT_MAX;
while(right<nums.size())
{
sum+=nums[right++];
while(sum>=s)
{
min1=min(min1,right-left);
sum-=nums[left++];
}
}
return min1==INT_MAX?0:min1;
}
};注：中等题，双指针法解决。faster than 49.08%。
210. Course Schedule II
注：中等题，双指针法解决。faster than 49.08%。
211. Add and Search Word - Data structure design
注：中等题，本题其实有个无比风骚的代码。正常的是用map做的。faster than 37.46%。
212. Word Search II
注：中等题，本题其实有个无比风骚的代码。正常的是用map做的。faster than 37.46%。
213. House Robber II
class Solution {
public:
int rob(vector<int>& nums) {
vector<int> a(nums.size(),-1);
vector<int> b(nums.size(),-1);
if(nums.size()==1)
return nums[0];
return max(digui(nums,a,0,nums.size()-1),digui(nums,b,1,nums.size()));
}
int digui(vector<int>& nums,vector<int>& a,int k,int p)
{
if(k>=p)
return 0;
if(a[k]!=-1)
return a[k];
a[k]=max(digui(nums,a,k+2,p)+nums[k],digui(nums,a,k+1,p));
return a[k];
}
};注：中等题，本题相比于198题多出了头尾的限制条件，需要在主函数中进行限制，要头不要尾或要尾不要头，这点我没想到。faster than 40.67%。
214. Shortest Palindrome
注：中等题，本题其实有个无比风骚的代码。正常的是用map做的。faster than 37.46%。
215. Kth Largest Element in an Array
风骚代码：
class Solution {
public:
int findKthLargest(vector<int>& nums, int k) {
sort(nums.begin(),nums.end());
return nums[nums.size()-k];
}
};正常代码：
class Solution {
public:
int findKthLargest(vector<int>& nums, int k) {
map<int,int> a;
for(int i=0;i<nums.size();i++)
a[nums[i]]++;
int j=0;
for(map<int,int>::iterator i=a.begin();i!=a.end();i++)
{
j+=i->second;
if(nums.size()-j<=k-1)
return i->first;
}
}
};注：中等题，本题其实有个无比风骚的代码。正常的是用map做的。faster than 37.46%。
216. Combination Sum III
class Solution {
public:
vector<vector<int>> combinationSum3(int k, int n) {
vector<vector<int>> a;
vector<int> b;
digui(a,b,k,n,0,1);
return a;
}
void digui(vector<vector<int>>& a,vector<int>& b,int k, int n,int p,int q)
{
if(b.size()==k)
{
if(p==n)
a.push_back(b);
return;
}
for(int i=q;i<=9;i++)
{
b.push_back(i);
digui(a,b,k,n,p+i,i+1);
b.pop_back();
}
}
};注：中等题，典型的回溯法。faster than 100%。
Rrui的Leetcode算法刷题笔记（五）链接如下：