LeetCode之maxProfitII

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解法一：

int maxprofit2(vector<int> &prices)  //Best Time to Buy and Sell Stock II {//cout<<prices[0];if (prices.size()==0||prices.size()==1){return 0;}int i=0,j=0,profit=0;while (prices[i]>=prices[i+1]){i++;if (i>=prices.size()-1){return 0;}}//i--;profit-=prices[i];while(prices[i]<=prices[i+1]) {i++;if (i>=prices.size()-1){return profit+=prices[i];}}profit+=prices[i];vector<int> sub_prices;for (j=i+1;j<=prices.size()-1;j++){sub_prices.push_back(prices[j]);}profit+=maxprofit2(sub_prices);return profit;}

解法二：

int maxprofit2(vector<int> &prices)  //Best Time to Buy and Sell Stock II {//cout<<prices[0];if (prices.size()==0||prices.size()==1){return 0;}int i=0,j=0,profit=0;while (prices[i]>=prices[i+1]){i++;if (i>=prices.size()-1){return 0;}}//i--;profit-=prices[i];while(prices[i]<=prices[i+1]) {i++;if (i>=prices.size()-1){return profit+=prices[i];}}profit+=prices[i];vector<int> sub_prices;for (j=i+1;j<=prices.size()-1;j++){sub_prices.push_back(prices[j]);}profit+=maxprofit2(sub_prices);return profit;}

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