LeetCode(123) Best Time to Buy and Sell Stock III
题目
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).
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分析
紧接着121 和122 的另外一道题目,此次要求只能进行两次买卖交易,求最大利润。
一篇很好的分析文章,参考博客。
第一步扫描,先计算出子序列[0,…,i]中的最大利润,用一个数组保存下来,那么时间是O(n)。
计算方法也是利用第一个问题的计算方法。
第二步是逆向扫描,计算子序列[i,…,n-1]上的最大利润,这一步同时就能结合上一步的结果计算最终的最大利润了,这一步也是O(n)。
第三步,求[0,i]的最大利润与[i,n-1]的最大利润之和的最大值,所以最后算法的复杂度就是O(n)的。
AC代码
class Solution {
public:
int maxProfit(vector<int>& prices) {
if (prices.empty())
return 0;
int n = prices.size();
vector<int> profits(n, 0), profits_reverse(n,0);
//正向遍历,profits[i]表示 prices[0...i]内做一次交易的最大收益.
int low = prices[0] , cur_profit = 0;
for (int i = 1; i < n; ++i)
{
if (prices[i] < low)
{
low = prices[i];
}
else{
if (cur_profit < prices[i] - low)
cur_profit = prices[i] - low;
}
profits[i] = cur_profit;
}//for
//逆向遍历, profits_reverse[i]表示 prices[i...n-1]内做一次交易的最大收益.
//当前最大价格
int high = prices[n - 1];
cur_profit = 0;
for (int i = n - 2; i >= 0; --i)
{
if (prices[i] > high)
high = prices[i];
else{
if (cur_profit < high - prices[i])
cur_profit = high - prices[i];
}//else
profits_reverse[i] = cur_profit;
}
int max_profile = 0;
for (int i = 0; i < n; i++)
{
if ((profits[i] + profits_reverse[i]) > max_profile)
max_profile = profits[i] + profits_reverse[i];
}
return max_profile;
}
};
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