Best Time to Buy and Sell Stock III
2014-07-28 13:45
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本解法启发自http://blog.csdn.net/pickless/article/details/12034365, 以下文字部分亦转自该文
O(n^2)的算法很容易想到:
找寻一个点j,将原来的price[0..n-1]分割为price[0..j]和price[j..n-1],分别求两段的最大profit。
进行优化:
对于点j+1,求price[0..j+1]的最大profit时,很多工作是重复的,在求price[0..j]的最大profit中已经做过了。
类似于Best Time to Buy and Sell Stock,可以在O(1)的时间从price[0..j]推出price[0..j+1]的最大profit。
但是如何从price[j..n-1]推出price[j+1..n-1]?反过来思考,我们可以用O(1)的时间由price[j+1..n-1]推出price[j..n-1]。
最终算法:
数组l[i]记录了price[0..i]的最大profit,
数组r[i]记录了price[i..n]的最大profit。
已知l[i],求l[i+1]是简单的,同样已知r[i],求r[i-1]也很容易。
最后,我们再用O(n)的时间找出最大的l[i]+r[i],即为题目所求。
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
class Solution
{
public:
//注释部分为 O(n^2)的解法
/*
int subProfit(vector<int> &p, int lhs, int rhs)
{
int len = rhs - lhs + 1;
int minv = p[lhs], profit = 0;
for(int i = lhs + 1; i <= rhs; ++i)
{
if(profit < p[i] - minv)
profit = p[i] - minv;
if(p[i] < minv)
minv = p[i];
}
return profit;
}
int maxProfit(vector<int> &prices)
{
int len = prices.size();
if(len < 2)
return 0;
int maxv = 0;
for(int i = 1; i < len - 1; ++i)
{
int tmp = subProfit(prices, 0, i) + subProfit(prices, i+1, len-1);
if(maxv < tmp)
maxv = tmp;
}
return maxv;
}*/
int maxProfit(vector<int> &prices)
{
int len = prices.size();
if(len < 2)
return 0;
int *l = new int[len]; //用以记录 prices[0 ~ i] 最大的收益
int *r = new int[len]; //用以记录 prices[i ~ len -1] 最大的收益
memset(l, 0, sizeof(int) * len);
memset(r, 0, sizeof(int) * len);
int minv = prices[0];
for(int i = 1; i < len; ++i)
{
l[i] = max(prices[i] - minv, l[i-1]);
minv = min(prices[i], minv);
}
int maxv = prices[len - 1];
for(int j = len - 2; j >= 0; --j)
{
r[j] = max(maxv - prices[j], r[j + 1]);
maxv = max(prices[j], maxv);
}
int re = 0;
for(int i = 0; i < len; ++i)
{
re = re > l[i] + r[i] ? re : l[i] + r[i];
}
delete[] l;
delete[] r;
return re;
}
};
int main(void)
{
/*
int a[] = {
3, 7, 8, 2, 9, 6, 1, 10, 5
};*/
//int a[] = {
// 4, 2, 1
// };
//
int a[] = {
1, 2, 4
};
int len = sizeof(a) / sizeof(int);
vector<int> v(a, a + len);
Solution s;
cout << s.maxProfit(v) << endl;
return 0;
}
O(n^2)的算法很容易想到:
找寻一个点j,将原来的price[0..n-1]分割为price[0..j]和price[j..n-1],分别求两段的最大profit。
进行优化:
对于点j+1,求price[0..j+1]的最大profit时,很多工作是重复的,在求price[0..j]的最大profit中已经做过了。
类似于Best Time to Buy and Sell Stock,可以在O(1)的时间从price[0..j]推出price[0..j+1]的最大profit。
但是如何从price[j..n-1]推出price[j+1..n-1]?反过来思考,我们可以用O(1)的时间由price[j+1..n-1]推出price[j..n-1]。
最终算法:
数组l[i]记录了price[0..i]的最大profit,
数组r[i]记录了price[i..n]的最大profit。
已知l[i],求l[i+1]是简单的,同样已知r[i],求r[i-1]也很容易。
最后,我们再用O(n)的时间找出最大的l[i]+r[i],即为题目所求。
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
class Solution
{
public:
//注释部分为 O(n^2)的解法
/*
int subProfit(vector<int> &p, int lhs, int rhs)
{
int len = rhs - lhs + 1;
int minv = p[lhs], profit = 0;
for(int i = lhs + 1; i <= rhs; ++i)
{
if(profit < p[i] - minv)
profit = p[i] - minv;
if(p[i] < minv)
minv = p[i];
}
return profit;
}
int maxProfit(vector<int> &prices)
{
int len = prices.size();
if(len < 2)
return 0;
int maxv = 0;
for(int i = 1; i < len - 1; ++i)
{
int tmp = subProfit(prices, 0, i) + subProfit(prices, i+1, len-1);
if(maxv < tmp)
maxv = tmp;
}
return maxv;
}*/
int maxProfit(vector<int> &prices)
{
int len = prices.size();
if(len < 2)
return 0;
int *l = new int[len]; //用以记录 prices[0 ~ i] 最大的收益
int *r = new int[len]; //用以记录 prices[i ~ len -1] 最大的收益
memset(l, 0, sizeof(int) * len);
memset(r, 0, sizeof(int) * len);
int minv = prices[0];
for(int i = 1; i < len; ++i)
{
l[i] = max(prices[i] - minv, l[i-1]);
minv = min(prices[i], minv);
}
int maxv = prices[len - 1];
for(int j = len - 2; j >= 0; --j)
{
r[j] = max(maxv - prices[j], r[j + 1]);
maxv = max(prices[j], maxv);
}
int re = 0;
for(int i = 0; i < len; ++i)
{
re = re > l[i] + r[i] ? re : l[i] + r[i];
}
delete[] l;
delete[] r;
return re;
}
};
int main(void)
{
/*
int a[] = {
3, 7, 8, 2, 9, 6, 1, 10, 5
};*/
//int a[] = {
// 4, 2, 1
// };
//
int a[] = {
1, 2, 4
};
int len = sizeof(a) / sizeof(int);
vector<int> v(a, a + len);
Solution s;
cout << s.maxProfit(v) << endl;
return 0;
}
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