标签:des c style class blog code
Problem Description:
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.
Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
按照题意,要求根据每天的股票价格,交易两次得到最大的利润。直接的想法就是利用分治法,从前往后循环依次将数组分为前后两个股票价格序列,分别得到最大的利润,将两者相加得到两次交易的最大利润,时间复杂度O(n^2),提交果然超时了。
代码如下:
class Solution {
public:
int profit(vector<int> &vec,int beg,int last)
{
int res=0;
if(beg==last)
return res;
int min=vec[beg];
for(int i=beg+1;i<=last;++i)
{
if(vec[i]>min)
{
int temp=vec[i]-min;
if(temp>res)
res=temp;
}
else
min=vec[i];
}
return res;
}
int maxProfit(vector<int> &prices) {
int res=0;
int n=prices.size();
if(n<=1)
return res;
for(int i=1;i<n-1;i++)
{
int res1=profit(prices,0,i);
int res2=profit(prices,i+1,n-1);
if((res1+res2)>res)
res=res1+res2;
}
return res;
}
};具体代码如下:
class Solution {
public:
int maxProfit(vector<int> &prices) {
int res=0;
int n=prices.size();
if(n<=1)
return res;
vector<int> front(n,0);
vector<int> back(n,0);
int minp=prices[0];
int maxp=prices[n-1];
for(int i=1;i<n;i++)
{
if(prices[i]>minp)
front[i]=prices[i]-minp;
else
minp=prices[i];
}
for(int i=n-2;i>=0;i--)
{
if(prices[i]>maxp)
maxp=prices[i];
back[i]=max(maxp-prices[i],back[i+1]);
if(front[i]+back[i]>res)
res=front[i]+back[i];
}
return res;
}
};Leetcode--Best Time to Buy and Sell Stock III,布布扣,bubuko.com
Leetcode--Best Time to Buy and Sell Stock III
标签:des c style class blog code
原文地址:http://blog.csdn.net/longhopefor/article/details/27340055