Leetcode123:best time to Buy and Sell Stock III

Say you has 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 are in most of the transactions.

Note:

Engage in multiple transactions on the same time (ie, you must sell the stock before you buy again).

Only two stock trading issues.

This problem can be converted to the best time to Buy and Sell Stock I issue.

The core of the two stock trades is the ability to define a trading point, which can be traded before the trading point (the maximum amount of money earned is firstprof), after which a trade can be made (the maximum amount of money earned is secondprof). Then Max (FIRSTPROF+SECONDPROF) is required. However, the time complexity of this method is O (n^2), and the spatial complexity is O (1). Timeout is displayed in Leetcode.

You can use the two-scan method to avoid the double loops above.

Unlike the initial state defined in the best time to Buy and Sell Stock I, A[i] represents the maximum amount of money sold in the first day, this further direct definition A[i] represents the maximum amount of money earned before I day. Minprice represents the lowest price from the No. 0 day to the I-1 day.

A[0]=0. (Initial state)

A[1]=max (Prices[1]-prices[0],a[0])

A[2]=max (Prices[2]-minprice,a[1])

.....

That is A[i]=max (Price[i]-minprice,a[i-1]).

A[0]=0

Another scan scans forward from the array, defining b[i] to represent the maximum amount of money that can be earned from day I to last day N. Maxprice represents the highest price for i+1 days to n days.

B[n]=0. (Initial state)

B[n-1]=max (Maxprice-prices[n-1],b[n])

B[n-2]=max (Maxprice-prices[n-2],b[n-1])

.....

That is B[i]=max (maxprice-prices[i],b[i+1])

B[n]=0

Then the maximum amount of money that can be earned by dividing points on day I is a[i]+b[i]

The solution for the problem is max{a[i]+b[i]}. 0<=i<=n.

The time complexity is O (n), and the spatial complexity is O (n).

Runtime:8ms

Class Solution {Public:int Maxprofit (vector<int>& prices) {int length=prices.size (); if (length==0| |        Length==1) return 0;        int * Ascandmax=new int[length] ();        int minprice=prices[0];        int maxprof=0;            for (int i=1;i<length;i++) {Maxprof=max (maxprof,prices[i]-minprice);            Minprice=min (Minprice,prices[i]);        Ascandmax[i]=maxprof;        } int* descandmax=new Int[length] ();        int maxprice=prices[length-1];        maxprof=0;            for (int i=length-2;i>=0;i--) {Maxprof=max (maxprof,maxprice-prices[i]);            Maxprice=max (Maxprice,prices[i]);        Descandmax[i]=maxprof;        } maxprof=0;        for (int i=0;i<length;i++) {Maxprof=max (maxprof,ascandmax[i]+descandmax[i]);        } delete [] Ascandmax;        delete [] Descandmax;    return maxprof; }};

Leetcode123:best time to Buy and Sell Stock III