leetcode_question_123 Best Time to Buy and Sell Stock III

Say you have an array for which the i^th element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete at most two transactions.

Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

和前两道题比起来，因为限制了2次交易，这道题最难了。

设i从0到n-1，那么针对每一个i，看看在prices的子序列[0,...,i][i,...,n-1]上分别取得的最大利润（第一题）即可。
这样初步一算，时间复杂度是O(n2)。

改进：
改进的方法就是动态规划，那就是第一步扫描，先计算出子序列[0,...,i]中的最大利润，用一个数组保存下来，那么时间是O(n)。
第二步是逆向扫描，计算子序列[i,...,n-1]上的最大利润，这一步同时就能结合上一步的结果计算最终的最大利润了，这一步也是O(n)。
所以最后算法的复杂度就是O(n)的。

int maxProfit(vector<int> &prices) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        int length = prices.size();    if(length < 2) return 0;//profit from 0 to iint* profitleft = new int[length];profitleft[0] = 0;int lowprice = prices[0];for(int i = 1; i < length; ++i){if(prices[i] > lowprice){int tmp = prices[i] - lowprice;if(tmp > profitleft[i-1])profitleft[i] = tmp;elseprofitleft[i] = profitleft[i-1];}else{lowprice = prices[i];profitleft[i] = profitleft[i-1];}}//profit from i to nint* profitright = new int[length];profitright[length-1] = 0;int highprice = prices[length-1];for(int i = length-2; i >= 0; --i){if(prices[i] < highprice){int tmp = highprice - prices[i];if(tmp > profitright[i+1])profitright[i] = tmp;elseprofitright[i] = profitright[i+1];}else{highprice = prices[i];profitright[i] = profitright[i+1];}}//int max = 0;for(int i = 0; i < length; ++i){int tmp = profitleft[i] + profitright[i];if(tmp > max) max = tmp;}return max;    }

int maxProfit(vector<int> &prices) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        int length = prices.size();    if(length < 2) return 0;//profit from 0 to iint* profitleft = new int[length];profitleft[0] = 0;int lowprice = prices[0];for(int i = 1; i < length; ++i){if(prices[i] > lowprice){int tmp = prices[i] - lowprice;if(tmp > profitleft[i-1])profitleft[i] = tmp;elseprofitleft[i] = profitleft[i-1];}else{lowprice = prices[i];profitleft[i] = profitleft[i-1];}}int max = profitleft[length-1];//profit from i to nint* profitright = new int[length];profitright[length-1] = 0;int highprice = prices[length-1];for(int i = length-2; i >= 0; --i){if(prices[i] < highprice){int tmp = highprice - prices[i];if(tmp > profitright[i+1])profitright[i] = tmp;elseprofitright[i] = profitright[i+1];}else{highprice = prices[i];profitright[i] = profitright[i+1];}int tmpmax = profitleft[i] + profitright[i];if(tmpmax > max) max = tmpmax;}return max;    }

精简代码：

class Solution {public:    int maxProfit(vector<int>& prices) {        int sz = prices.size();        if(sz < 2) return 0;        vector<int> lprofit(sz, 0);        vector<int> rprofit(sz, 0);        int lmin = prices[0], rmax = prices[sz-1];        for(int i = 1; i < sz; i++){            lprofit[i] = max(prices[i]-lmin, lprofit[i-1]);            if(prices[i] < lmin) lmin = prices[i];                        rprofit[sz-i-1] = max(rmax-prices[sz-i], rprofit[sz-i]);            if(prices[sz-i-1] > rmax) rmax = prices[sz-i-1];        }        int profit = 0;        for(int i=0; i <sz;i++){            profit = max(profit, lprofit[i]+rprofit[i]);        }        return profit;    }};