121. Best Time to Buy and Sell Stock
2018-02-09 16:44
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为求最大子序列和问题 53题
nums[i] = prices[i]-prices[i-1]
nums[0] = 0;
为求最大子序列和问题 53题
nums[i] = prices[i]-prices[i-1]
nums[0] = 0;
#include <cstdio> #include <iostream> #include <set> #include <vector> using namespace std; int maxProfit(vector<int>& prices) { int sum = 0; int maxsum = 0; for (int i=1;i<prices.size();++i) { if (sum > 0) sum += prices[i]-prices[i-1]; else sum = prices[i]-prices[i-1]; if (sum >maxsum) maxsum = sum; } return maxsum; } int main() { vector<int> a; int x; for(int i=0;i<6;i++) { cin>>x; a.push_back(x); } /*int val; cin>>val; /*for ( int ix = 0; ix < a.size(); ++ix ) cout << a[ ix ] <<endl;*/ cout<<maxProfit(a)<<endl; return 0 ; }
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