#123 Best Time to Buy and Sell Stock III

Posted 2020-10-13 东方春

/*
找一次操作能获取最大利润的一次交易（即题目121）的idxBuy和idxSell，这次交易称为首次交易
然后分5中情况：
情况1：最优解就只有“首次交易”
情况2：“首次交易”加上左边的交易得到最优解
情况3：“首次交易”加上右边的交易得到最优解
情况4：“首次交易”不做，两边的交易得到最优解
情况5：“首次交易”不做，idxBuy和idxSell中间找亏损最大的交易idxBuy2,idxSell2，然后idxBuy买,idxBuy2卖，idxSell2买，idxSell卖就是最优解
*/
class Solution {
public:
    #define MAX(a,b) ((a)>(b)?(a):(b))
    
	int maxProfit(vector<int>& prices) {
        if(prices.size() < 2)return 0;
        
        //处理首次交易
        int idxBuy = 0 ,idxSell = 0;
        int profitOrigin = this->maxProfitOfOne(prices, 0, prices.size()-1,&idxBuy,&idxSell);

        int maxProfit = profitOrigin;//默认

        //算出左侧交易利润
        int profitLeft = this->maxProfitOfOne(prices, 0, idxBuy-1);
        
        //算出右侧交易利润
        int profitRight = this->maxProfitOfOne(prices, idxSell+1, prices.size()-1);
        
        //算出情况1,2,3,4的最大利润
        maxProfit = MAX(profitLeft + profitOrigin, profitRight + profitOrigin);
        maxProfit = MAX(maxProfit,profitLeft + profitRight);
        
        //算出idxBuy，idxSell之间的最大亏损,反过来操作就是更大的盈利
        vector<int> pricesRe = prices;
        reverse(pricesRe.begin(),pricesRe.end());
        
        int maxProfitEx = this->maxProfitOfOne(pricesRe, pricesRe.size() -1- idxSell, pricesRe.size() -1- idxBuy);
        
        return MAX(maxProfitEx+profitOrigin,maxProfit);
        
    }


	int maxProfitOfOne(vector<int>& prices, int left, int right,int *pMin=NULL,int *pMax=NULL) {
		if (left >= right)
			return 0;

		int minLop = left, maxLop = left;
		int maxProfit = 0;

		int idx = minLop + 1;
		while (idx < right + 1) {
			if (prices[idx] > prices[maxLop]) {//find larger
				maxLop = idx;
				int profit = prices[maxLop] - prices[minLop];
				if (profit > maxProfit) {
					maxProfit = profit;
                    if(pMin)*pMin = minLop;
                    if(pMax)*pMax = maxLop;
				}
			}
			else if (prices[idx] < prices[minLop]) {//find smaller
				minLop = idx;
				maxLop = idx;
			}
			idx++;
		}

		return maxProfit;
	}

};