11. 盛最多水的容器
class Solution {
public:
int maxArea(vector<int>& height) {
vector<int> & v =height;
int l =0,r=v.size()-1;
int res = 0;
while(l<r)
{
int lval = v[l];
int rval = v[r];
int minval = min(lval,rval);
res =max(res,minval*(r-l));
if(lval<rval) l++;
else r--;
}
return res ;
}
};
34. 搜索范围
class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
vector<int> v(2,-1);
int high =binarySearchUp(nums,target,0,nums.size()-1);
int low =binarySearchDown(nums,target,0,nums.size()-1);
//DBG2(high,low);
if(high >= low )
{
v[0]=low;
v[1]=high;
}
return v;
}
private:
int binarySearchUp(vector<int> & nums ,int target,int begin,int end)
{
if(begin>end)return end;
int mid =(begin+end) >>1;
if(nums[mid]>target) return binarySearchUp(nums,target,begin,mid-1);
else return binarySearchUp(nums,target,mid+1,end);
}
int binarySearchDown(vector<int> & nums ,int target,int begin,int end)
{
if(begin>end)return begin;
int mid =(begin+end) >>1;
if(nums[mid]<target) return binarySearchDown(nums,target,mid+1,end);
else return binarySearchDown(nums,target,begin,mid-1);
}
};
59. 螺旋矩阵 II
class Solution {
public:
vector<vector<int> > generateMatrix(int n) {
vector<vector<int> > v (n,vector<int> (n,0));
int col_min=0,col_max=n-1,row_min=0,row_max=n-1;
int row=-1,col=0;
int i=1;
while(i<=n*n)
{
while(++row<=row_max){
v[col][row]=i++;
}
row--;
while(++col<=col_max) {
v[col][row]=i++;
}
col --; row_max--;
while(--row>=row_min) {
v[col][row]=i++;
} row++; col_max--; col_min++;
while(--col>=col_min) {
v[col][row]=i++;
} col++;row_min++;
}
return v;
}
};
60. 第k个排列
class Solution {
public:
string getPermutation(int n, int k) {
vector<int> v;
for(int i=1;i<=n;i++) v.push_back(i);
for(int i=0;i<k;i++)
{
next_permutation(v.begin(),v.end());
};
string str ="";
for(auto i :v)
{
str+=i;
}
return str;
}
};
141. 环形链表
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
bool hasCycle(ListNode *head) {
if(head==NULL) return false;
ListNode * fast=head,*last=head;
if(fast->next&& fast->next->next) fast =fast->next->next;
else return false ;
if(last->next)last=last->next;
while(fast!=last)
{
if(fast->next&& fast->next->next) fast =fast->next->next;
else return false ;
if(last->next)last=last->next;
}
return true;
}
};
155. 最小栈
class MinStack {
public:
/** initialize your data structure here. */
stack<int>all;
stack<int>minstack;
MinStack() {
}
void push(int x) {
if(all.empty())
{
all.push(x);
minstack.push(x);
}
else
{
all.push(x);
if(minstack.top()>=x)minstack.push(x);;
}
}
void pop() {
if(all.top()==minstack.top()) minstack.pop();
all.pop();
}
int top() {
return all.top();
}
int getMin() {
return minstack.top();
}
};
/**
* 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();
*/
416. Partition Equal Subset Sum
class Solution {
public:
bool canPartition(vector<int>& nums) {
int sum = accumulate(nums.begin(), nums.end(), 0), target = sum >> 1;
if (sum & 1) return false;
vector<bool> dp(target + 1, false);
dp[0] = true;
for (int num : nums) {
for (int i = target ; i >=num ; --i) {
dp[i] = dp[i] || dp[i - num];
}
}
return dp[target];
}
};
374. 猜数字大小
// Forward declaration of guess API.
// @param num, your guess
// @return -1 if my number is lower, 1 if my number is higher, otherwise return 0
int guess(int num);
class Solution {
public:
int guessNumber(int n) {
int l=1,r=n;
int mid,ret;
while(l<r)
{
mid = l/2+r/2;
// mid=l+r>>1;
ret = guess(mid);
if(ret==0) return mid;
else if(ret< 0 ) r=mid-1;
else l=mid+1;
}
return l;
}
};
378. 有序矩阵中第K小的元素
class Solution {
public:
int kthSmallest(vector<vector<int>>& matrix, int k) {
priority_queue<int> q ;
for(auto & v : matrix) for(auto i:v)
{
q.push(i);
if(q.size()>k)q.pop();
} return q.top();
}
};
383. 赎金信
class Solution {
public:
bool canConstruct(string ransomNote, string magazine) {
// unordered_map<char,int> a;
int a[256] ={0};
for(auto i:magazine)
{
a[i]++;
}
for(auto i:ransomNote)
{
if(a[i]> 0 ) a[i]--;
else return false;
}
return true;
}
};
395. 至少有K个重复字符的最长子串
515. 在每个树行中找最大值
/**
* 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> largestValues(TreeNode* root) {
vector<int> v;
if(root==NULL) return v;
queue<TreeNode*> q ;
q.push(root);
while(1)
{
int tmp = getnum(q);
v.push_back(tmp);
getnxt(q) ;
if(q.empty())break;
}
return v;
}
void getnxt(queue<TreeNode*>& tmp )
{
queue<TreeNode*> t ;
while(!tmp.empty())
{
TreeNode* i=tmp.front();
if(i->left)
{
t.push(i->left);
}
if(i->right)
{
t.push(i->right);
}
tmp.pop();
}
while(!t.empty()) tmp.push(t.front()),t.pop();
}
int getnum(queue<TreeNode*> tmp)
{
int mm = -2147483648;
while(!tmp.empty())
{
int tt = tmp.front()->val;
mm=max(mm,tt);
tmp.pop();
}
return mm;
}
};
521. 最长特殊序列 Ⅰ
class Solution {
public:
int findLUSlength(string a, string b) {
return a==b?-1:max(a.size(),b.size());
}
};
540. 有序数组中的单一元素
class Solution {
public:
int singleNonDuplicate(vector<int>& nums) {
int ans =0;
for(auto i:nums)
ans^=i;
return ans;
}
};
639. 解码方法 2
class Solution {
public:
int ways(char c) {
if (c == ‘*‘)
return 9;
if (c == ‘0‘)
return 0;
return 1;
}
int ways(char i, char j) {
if (i == ‘*‘ && j == ‘*‘)
return 15;
if (i == ‘*‘)
return (j <= ‘6‘ ? 2 : 1);
if (j == ‘*‘) {
if (i == ‘1‘)
return 9;
if (i == ‘2‘)
return 6;
return 0;
}
if (i == ‘1‘ || (i == ‘2‘ && j <= ‘6‘))
return 1;
return 0;
}
int numDecodings(string s) {
int n =s.length();
if(n==0||s[0]==‘0‘) return 0;
long f2=1,f1= (long)ways(s[0]);
long res = f1;
long mod = (long)1e9+7;
for(int i=1;i<n;i++)
{
res = ways(s[i-1],s[i])*f2 + ways(s[i])*f1;
res%=mod;
f2=f1;
f1= res;
}
return res;
}
};
677. 键值映射
class MapSum {
public:
/** Initialize your data structure here. */
map<string,int> m;
MapSum() {
}
void insert(string key, int val) {
m[key] =val;
}
int sum(string prefix) {
int s = 0;
for (auto one : m)
{
if (prefix == one.first.substr(0, prefix.length()))
s += one.second;
}
return s;
}
};
/**
* Your MapSum object will be instantiated and called as such:
* MapSum obj = new MapSum();
* obj.insert(key,val);
* int param_2 = obj.sum(prefix);
*/
743. 网络延迟时间
#define REP(i,l,r) for(int i=l;i<=r;i++)
class Solution {
public:
int networkDelayTime(vector<vector<int>>& times, int N, int K) {
const int inf = (int)1e5+9;
int n=N,k=K;
int m[n+1][n+1];
int dist[n+1],vis[n+1];
REP(i,1,n) REP(j,1,n) if(i==j)m[i][j] ;else m[i][j]=inf;
for(auto i:times)m[i[0]][i[1]]=i[2];
REP(i,1,n) dist[i]=m[k][i],vis[i]=0;
vis[k]=1;
int ans =-1;
REP(i,1,n-1)
{
int minx=inf,u=k;
REP(j,1,n) if(!vis[j] && dist[j]!=inf &&dist[j]<minx) minx=dist[j],u=j;
if(minx==inf) return -1;
vis[u]=1;
ans=max(ans,minx);
REP(j,1,n) if(!vis[j]&& dist[j]>dist[u]+m[u][j])
dist[j]=dist[u]+m[u][j];
}
return ans;
}
};
原文地址：https://www.cnblogs.com/corx/p/9001830.html
时间： 2024-10-29 20:58:39