LeetCode OJ--Best Time to Buy and Sell Stock III
http://oj.leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/
这三道题,很好的进阶。1题简单处理,2题使用贪心,3题使用动态规划。
话说这叫一维动态规划,嗯。又复习了《算法导论》中和贪心以及动态规划有关的知识,记录如下:
动态规划的标志:最优子结构、子问题重叠。
1.找最优子结构
2.定义递归公示(列一个式子出来,并定义好这个式子到底是什么意思)。
3.写自底向上或递归备忘录法。
比如本问题:f(i,j) = max{f(i,k)+f(k,j)} 其中:f(i,j)表示从i到j的所有数,进行一次交易能获得的最多的钱数。最终要求的是f(0,n-1),k从1到n-2.
再深化一下:
fp(i,j)表示从i 到 j 进行最多进行 p 次交易可以获得的钱数,则:
fp(i,j) = max {fp-1(i,j), fp-p'(i,k)+fp'(k,j)} 其中:p'从1到p-1,k从i+1到j-1.
这大概叫二维动态规划(或许)。
贪心算法是从局部最优,能够做到全局最优。可分成一步步的,当前的选择受之前做过的选择的影响。而动态规划,是受后面要做的选择的影响。
于是有了下面代码,复杂度O(n2)然后超时了。
class Solution {
public:
//start 第一个元素,end 最后一个元素的下标
int onceBuyAndSell(vector<int> &prices,int start,int end)
{
int ans = ;
int max = prices[end];
for(int i = end - ;i>=start;i--)
{
if(max - prices[i]>ans)
ans = max - prices[i];
if(prices[i]>max)
max = prices[i];
}
return ans;
} int maxProfit(vector<int> &prices) {
if(prices.empty())
return NULL;
if(prices.size()==)
return ; int ans = onceBuyAndSell(prices,,prices.size()-); if(prices.size()< || ans == )
return ans; int temp = ;
for(int itor = ;itor<prices.size()-;itor++)
{
temp = onceBuyAndSell(prices,,itor) + onceBuyAndSell(prices,itor+,prices.size()-);
if(temp> ans)
ans = temp;
}
return ans;
}
};
使用了动态规划中打表的方法,降低了复杂度O(n).代码如下:
class Solution {
public:
vector<int> ansN;
//ansN[i]表示,从i到最后的所有元素,只进行一次交易的话,可以得到的最多值。
void onceBuyAndSell(vector<int> &prices,int start,int end)
{
int ans = ;
int max = prices[end];
ansN[end] = ;
for(int i = end - ;i>=start;i--)
{
if(max - prices[i]>ans)
ans = max - prices[i];
if(prices[i]>max)
max = prices[i];
ansN[i] = ans;
}
} int maxProfit(vector<int> &prices) {
if(prices.empty())
return NULL;
if(prices.size()==)
return ; ansN.resize(prices.size());
//先来一遍从后面计算的。
onceBuyAndSell(prices,,prices.size()-); if(prices.size()< )
return ansN[]; int min = prices[];
int ans2 = ; int ans = ansN[]; for(int itor = ;itor<prices.size()-;itor++)
{
if(prices[itor] - min>ans2)
ans2 = prices[itor] - min;
if(prices[itor]<min)
min = prices[itor]; if(ans2 + ansN[itor+]> ans)
ans = ans2 + ansN[itor+];
}
return ans;
}
};
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