[LeetCode] Best Time to Buy and keep Stock IV, leetcode=

[LeetCode] Best Time to Buy and keep Stock IV, leetcode=

Say you have an array for which 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 k transactions.

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

Solutions

Use Dynamic Planning to solve the problem.
We need to maintain the following two quantities:
global[i][j]: Currently reachesiDays at mostjThe maximum profit of the transaction.
local[i][j]: Currently reachesiDays at mostjTransaction, and the last transaction was sold on the day, the maximum profit.
State transition equation:
global[i][j] = max(local[i][j], global[i-1][j])
The above equation compares the values of two quantities: ① the current local maximum value; ② the past global maximum value.
local[i][j] = max(global[i-1][j-1] + max(diff, 0), local[i-1][j] + diff)
The above equation compares the two quantities:
① Globali-1Dayj-1Next, add the current transaction (if today's transaction makes money ).
② Obtain the local numberi-1DayjNext transaction, and then add the difference (local[i-1][j]Yesi-1Day-to-day transaction, which addsdiffAfteriSales per day does not increase the number of transactions. RegardlessdiffBoth positive and negative values must be added; otherwise, they cannot be satisfied.local[i][j]The condition that must be sold on the last day)

Another issue that needs to be noted is that when k is much larger than the array size, the above algorithm will become inefficient. Therefore, we can change it to an unlimited number of transactions.

Implementation Code

/*************************************** * ************************** @ Author: chu Xing * @ Date: 2015/2/22 * @ Status: Accepted * @ Runtime: 15 ms ************************************** * **************************/# include <iostream> # include <vector> # include <algorithm> using namespace std; class Solution {public: int maxProfit (int k, vector <int> & prices) {int len = prices. size (); if (len = 0) {return 0;} if (k> = len) {return helper (prices );} vector <int> max_local (k + 1, 0); vector <int> max_global (k + 1, 0); int diff; for (int I = 0; I <len-1; I ++) {diff = prices [I + 1]-prices [I]; for (int j = k; j> = 1; j --) {max_local [j] = max (max_global [j-1] + max (diff, 0), max_local [j] + diff ); max_global [j] = max (max_local [j], max_global [j]);} return max_global [k];} int helper (vector <int> & prices) {int profit = 0; for (int I = 0; I <prices. size ()-1; I ++) {profit = max (profit, profit + prices [I + 1]-prices [I]);} return profit ;}};