远景笔试题两道有先出来的大佬说了leetcode原题，贴代码
第一题股票最大亏损
任意次买卖股票，寻找当前峰值和低谷值，之前做的利润最大，只要将原来的数组逆序就可以了：
#include <iostream>
#include <vector>
#include <algorithm>
#include <sstream>
using namespace std;
int MaxKui(vector<int>& prices) {
if(prices.empty()) return 0;
int i = 0;
int peak = prices[0];
int valley = prices[0];
int profit = 0;
while(i < prices.size() - 1)
{
while(i < prices.size() - 1 && prices[i + 1] <= prices[i])
{
i++;
}
valley = prices[i];
while(i < prices.size() - 1 && prices[i + 1] > prices[i])
{
i++;
}
peak = prices[i];
profit += (peak - valley);
}
return profit;
}
int main()
{
string str="";
cin>>str;
str+=',';
istringstream is(str);
vector<int> arr;
int tmp;
char c;
while(is>>tmp>>c)
{
arr.emplace_back(tmp);
}
reverse(arr.begin(),arr.end());
if(MaxKui(arr)<0)
{
cout<<0<<endl;
}else
cout<<MaxKui(arr)<<endl;
return 0;
} 第二题最大光伏面板群：dfs遍历，遍历过的点清0：
#include <iostream>
#include <vector>
#include <cmath>
#include <string>
#include <sstream>
using namespace std;
int dfs(int i,int j,vector<vector<int> > &grid )
{
int m=grid.size();
int n=grid[0].size();
int area=1;
grid[i][j]=0;
if(i-1>=0&&grid[i-1][j]==1)area+=dfs(i-1,j,grid);
if(i+1<m&&grid[i+1][j]==1)area+=dfs(i+1,j,grid);
if(j-1>=0&&grid[i][j-1]==1)area+=dfs(i,j-1,grid);
if(j+1<n&&grid[i][j+1]==1)area+=dfs(i,j+1,grid);
return area;
}
int maxArea(vector<vector<int> > &grid)
{
int m=grid.size();
int n=grid[0].size();
if(m==0)
return 0;
int ans=0;
for(int i=0;i<m;++i)
{
for(int j=0;j<n;++j)
{
if(grid[i][j]==1)
{
ans=max(ans,dfs(i,j,grid));
}
}
}
return ans;
}
int main()
{
int n,m;
char c;
cin>>n>>c>>m;
vector<vector<int> >grid;
for(int i=0;i<n;++i)
{
string str;
int temp;
char c;
cin>>str;
vector<int>arr;
str+=',';
istringstream is(str);
while(is>>temp>>c)
{
arr.emplace_back(temp);
}
grid.emplace_back(arr);
}
cout<<maxArea(grid)<<endl;
return 0;
} 今天错过科大讯飞的笔试，远景笔试过了给面试吧。。。。。
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