Best Time to Buy and Sell Stock III O(n) 求解方法

leetcode的题目：http://oj.leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/

leetcode的题目都很简练，但是很有趣，认真地做吧。

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).

a. 对于输入为n，最多可以买卖两次，求最大的盈利。

b. 买两次的收益通常比买一次来的大。

c. 对于输入规模为n，当n>=4时可以将n分为两部分，且分类的方法有n-3种。每份的最少个数为2。因为n个元素，总共有n-1条缝隙可以划分，又最左或最右至少要有两个元素，所以n-1-2 = n-3 种。

d. 求得划分一次（买卖两次）的每种划分的最大盈利，和不划分的最大盈利，即得到最大的盈利。

　　max_profit = max{买卖两次（划分一次），买卖一次（不进行划分）}

e. 定义状态矩阵dp_be[i], 表示以0为起点，i为终点的子串的最大盈利。

   dp_be[0] = 0;int i, max_profit = 0, cur_min = prices[0];for(i = 1; i < len; i++) {max_profit = max(max_profit, prices[i]-cur_min);cur_min = min(cur_min, prices[i]);dp_be[i] = max_profit; // 在i时卖出的最大收益}

f. 定义状态矩阵dp_af[i],表示以i为起点，n-1(最后元素)为终点的子串的最大盈利

 max_profit = 0; int cur_max = prices[len-1];for(i = len-2; i >= 0; i--) {max_profit = max(max_profit, cur_max-prices[i]);cur_max = max(cur_max, prices[i]);dp_af[i] = max_profit;}

g. 由此可得，对于以i为划分点，将n个元素分成两个子串，要求两个子串之和最大，根据贪心算法思想，要求每个子串都去的最大值。对i从1到n-3循环得到最大值。

     // 放入只买一次的最大值int max = dp_be[len-1];// 在同一天卖了再买是没有意义的// 求两个数组和的最大值for(i = 1; i < len-1; i++) {if(dp_be[i]+dp_af[i+1] > max)max = dp_be[i]+dp_af[i+1];}

主要：dp_be[i]表示0开始，i结束的子串的最大盈利，所以它对应的尾部子串为dp_af[i+1]，表示以i+1开始，n-1为终点的子串的最大盈利。

h. 代码

int Solution::maxProfit2(vector<int> &prices) {int len = prices.size();if(len < 2)return 0;int *dp_be = new int[len]; // 记录前向最大int *dp_af = new int[len]; // 记录后向最大dp_be[0] = 0;int i, max_profit = 0, cur_min = prices[0];for(i = 1; i < len; i++) {max_profit = max(max_profit, prices[i]-cur_min);cur_min = min(cur_min, prices[i]);dp_be[i] = max_profit; // 在i时卖出的最大收益}dp_af[len-1] = 0;max_profit = 0; int cur_max = prices[len-1];for(i = len-3; i >= 0; i--) {max_profit = max(max_profit, cur_max-prices[i]);cur_max = max(cur_max, prices[i]);dp_af[i] = max_profit;}if(len == 2)return dp_be[len-1];else {// 放入只买一次的最大值int max = dp_be[len-1];// 在同一天卖了再买是没有意义的// 求两个数组和的最大值for(i = 1; i < len-1; i++) {if(dp_be[i]+dp_af[i+1] > max)max = dp_be[i]+dp_af[i+1];}return max;}
}

i. 画个图说明以下

转载于:https://www.cnblogs.com/themo/p/3684113.html

Best Time to Buy and Sell Stock III O(n) 求解方法相关推荐

【DP + 卖股票】LeetCode 123. Best Time to Buy and Sell Stock III
LeetCode 123. Best Time to Buy and Sell Stock III Solution1: 不得不让人感叹算法之精妙啊!!! 参考网址:[1]http://www.cnb ...
Leetcode Best Time to Buy and Sell Stock III
Leetcode Best Time to Buy and Sell Stock III,本算法的关键为找出其动态子结构.可以发现,序列中的最小值可以做为其的一个分割,令左边序列为left,右边的序列 ...
[LeetCOde][Java] Best Time to Buy and Sell Stock III
题目: Say you have an array for which the ith element is the price of a given stock on day i. Design a ...
Leetcode_123_Best Time to Buy and Sell Stock III
本文是在学习中的总结,欢迎转载但请注明出处:http://blog.csdn.net/pistolove/article/details/43740415 Say you have an array ...
Best Time to Buy and Sell Stock III
解题思路来自:https://blog.csdn.net/u012501459/article/details/46514309 Say you have an array for which the ...
最多两次股票交易-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 al ...
LeetCode OJ - Best Time to Buy and Sell Stock III
题目: Say you have an array for which the ith element is the price of a given stock on day i. Design a ...
【Best Time to Buy and Sell Stock III 】cpp
题目: Say you have an array for which the ith element is the price of a given stock on day i. Design a ...
leetcode 123. Best Time to Buy and Sell Stock III | 123. 买卖股票的最佳时机 III（总结DP 模型套路）
题目 https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/description/ DP 模型套路 DP 套路之:暴力递 ...

Best Time to Buy and Sell Stock III O(n) 求解方法

Best Time to Buy and Sell Stock III O(n) 求解方法相关推荐

最新文章

热门文章