Best Time to Buy and Sell Stock I II III

I

对于第一题，本质意义是求出数组中某两个元素（day[i],day[i]）的最大差值。但是，要求j>i

int maxProfit(vector<int> &prices) {        if(prices.size() ==0)            return 0;                int min=prices[0], profit=0;        for(int i=0; i<prices.size(); i++)        {            if(prices[i] - min > profit)                profit = prices[i] - min;                            if(prices[i] < min)                min = prices[i];                                }                return profit;    }

II

第二题是可以进行任意多次交易，那么对数组遍历，找到临时最小值和临时最大值。

int maxProfit(vector<int> &prices) {        if (prices.size() == 0)return 0;int profit = 0, tmpMin=prices[0], tmpMax=prices[0];int n = prices.size();for (int i = 1; i < n; i++){    //寻找比tmpMIN小的元素if (prices[i] < tmpMin){    //比tmpMin小，则更新较小的tmpMin = prices[i];}else{    //比tmpMin大，则可以出手卖掉，并更新tmpMinprofit += (prices[i]-tmpMin);tmpMin = prices[i];}}return profit;    }

III

第三题比较复杂，只允许买卖两次，有两种方法。

其一，由于两次操作之间没有交互，即第一次买卖完成之后才能进行第二次的买卖，因此可以用分治法将[0...n]中取任意一个i，分别去求[0...i]和[i+1...n-1]的最大利润，那么两者之和就是两次操作的最大利润，但是这个算法的时间复杂度是O^2，超时了。

//取[m, n]之间的最大利润int Profit1(vector<int>& prices, int m, int n){int size = prices.size();if(m >= size)return 0;int minPrice = prices[m], profit = 0;for (int i = m+1; i < n; i++){if (prices[i] - minPrice > profit){profit = prices[i] - minPrice;}if (prices[i] < minPrice){minPrice = prices[i];}}return profit;}int maxProfit(vector<int> &prices){if(prices.size() == 0)return 0;int n = prices.size();int max1 = 0;for (int i = 1; i < n; i++){int tmp1 = Profit1(prices, 0, i);int tmp2 = Profit1(prices, i+1, n);if(tmp1 + tmp2 > max1)max1 = tmp1 + tmp2;}return max1;}

方法二，DP，参考了一位网友的方法：http://blog.csdn.net/pickless/article/details/12034365

DP的关键是找出地推关系式

int maxProfit(vector<int> &prices){int n = prices.size();if(0==n || 1==n)return 0;int * l = new int[n];int * r = new int[n];memset(l, 0, sizeof(int) * n);memset(r, 0, sizeof(int) * n);int minPrice = prices[0];for (int i = 1; i < n; i++){l[i] = prices[i] - minPrice > l[i-1]? prices[i] - minPrice : l[i-1];minPrice = prices[i] < minPrice ? prices[i] : minPrice;}int maxPrice = prices[n-1];for (int i = n-2; i >=0; i--){r[i] = maxPrice - prices[i] > r[i+1] ? maxPrice - prices[i] : r[i+1];maxPrice = prices[i] > maxPrice ? prices[i] : maxPrice;}int profit = 0;for (int i = 0; i < n; i++){profit = l[i] + r[i] > profit ? l[i] + r[i] : profit;}delete l;delete r;return profit;}