[leetcode] Best Time to Buy and Sell Stock III

from : https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/

Say you have an array for which the ith 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.

class Solution {public:    int maxProfit(vector<int>& prices) {        int days = prices.size();        if(days <= 1) {            return 0;        }        int *left = new int[days]; // left[i]:inlcude ith day, can make most.        int *right = new int[days];                // for left        int v = prices[0];        left[0] = 0;        for(int i=1; i<days; ++i) {            if(v > prices[i]) {                v = prices[i];            }            left[i] = max(left[i-1], prices[i] - v);        }                // for right        v = prices[days-1];        right[days-1] = 0;        for(int i=days-2; i>=0; --i) {            if(v < prices[i]) {                v = prices[i];            }            right[i] = max(right[i+1], v - prices[i]);        }                int made = 0;        for(int i=0; i<days; ++i) {            if(made < left[i] + right[i]) {                made = left[i] + right[i];            }        }                delete[] left;        delete[] right;        return made;    }};

public class Solution {    public int maxProfit(int[] prices) {        if(prices == null || prices.length < 2) {            return 0;        }        int n = prices.length;        int[] t1 = new int[n];// t1[i]：在i和i之前卖，最大获利        int[] t2 = new int[n];// t2[i]：在i和i之后买，最大获利                int a = prices[0];        int p = 0;        for(int i=1; i<n; ++i) {            if(prices[i] < a) {                a = prices[i];            }            if(prices[i] -a >p) {                p = prices[i]-a;            }            t1[i] = p;        }                a = prices[n-1];        p = 0;        for(int i=n-2; i>=0; --i) {            if(prices[i] > a) {                a = prices[i];            }            if(a - prices[i] > p) {                p = a-prices[i];            }            t2[i] = p;        }                p = 0;        for(int i=0; i<n; ++i) {            if(t1[i]+t2[i] > p) {                p = t1[i]+t2[i];            }        }        return p;    }}

最近被问如何求出买入和卖出点的问题，于是用java写了下面的代码。

package dp;import java.util.ArrayList;import java.util.List;public class StockBaidu {class Choice {int buyAt;int selAt;public Choice(int buy, int sel) {buyAt = buy;selAt = sel;} }class Profit {Choice chioce;int sum;public Profit(Choice c, int s) {chioce = c;sum = s;}}public List<Choice> maxProfit(int[] prices) {List<Choice> ans = new ArrayList<Choice>();int days = prices.length;if (days <= 1) {return ans;}Profit[] left = new Profit[days]; // left[i]:inlcude ith day, can make most.Profit[] right = new Profit[days];// for leftint k = 0;int v = prices[0];left[0] = new Profit(new Choice(0, 0),  0);for (int i = 1; i < days; ++i) {if (v > prices[i]) {v = prices[i];k = i;}left[i] = choose(left[i-1], prices[i]-v,  k, i);}// for rightk = days - 1;v = prices[k];right[days - 1] = new Profit(new Choice(days - 1, days - 1),  0);for (int i = days - 2; i >= 0; --i) {if (v < prices[i]) {v = prices[i];k = i;}right[i] = choose(right[i+1], v-prices[i], i, k);}int made = 0;for (int i = 0; i < days; ++i) {if (made < left[i].sum + right[i].sum) {ans.clear();if(left[i].chioce.buyAt < left[i].chioce.selAt) {ans.add(left[i].chioce);}if(right[i].chioce.buyAt < right[i].chioce.selAt) {ans.add(right[i].chioce);}made = left[i].sum + right[i].sum;}}return ans;}public int maximumProfit(int[] prices) {        int days = prices.length;        if(days <= 1) {            return 0;        }        int[] left = new int[days];        int[] right = new int[days];                // for left        int v = prices[0];        left[0] = 0;        for(int i=1; i<days; ++i) {            if(v > prices[i]) {                v = prices[i];            }            left[i] = Math.max(left[i-1], prices[i] - v);        }                // for right        v = prices[days-1];        right[days-1] = 0;        for(int i=days-2; i>=0; --i) {            if(v < prices[i]) {                v = prices[i];            }            right[i] = Math.max(right[i+1], v - prices[i]);        }                int made = 0;        for(int i=0; i<days; ++i) {            if(made < left[i] + right[i]) {                made = left[i] + right[i];            }        }                return made;    }private Profit choose(Profit before, int profit, int buyAt, int selAt) {if(before.sum >= profit) {return before;}return new Profit(new Choice(buyAt, selAt), profit);}public static void main(String[] args) {int[] data = {4,1,9,3,6,5,2,1,4,9,3,4,2,5,9};int profit = 0;for(Choice c : new StockBaidu().maxProfit(data)) {profit += data[c.selAt] - data[c.buyAt];System.out.println("profit is " + profit);System.out.println("buy at " + c.buyAt + ": sell at " + c.selAt);}System.out.println(new StockBaidu().maximumProfit(data));int[] data2 = {9, 2, 6};int profit2 = 0;for(Choice c : new StockBaidu().maxProfit(data2)) {profit2 += data2[c.selAt] - data2[c.buyAt];System.out.println("profit is " + profit2);System.out.println("buy at " + c.buyAt + ": sell at " + c.selAt);}System.out.println(new StockBaidu().maximumProfit(data2));}}