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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假設(shè)你有一個數(shù)組翩腐，它的第i個元素是一支給定的股票在第i天的價格俯抖。設(shè)計一個算法來找到最大的利潤愚臀。你最多可以完成兩筆交易范删。
注意事項
你不可以同時參與多筆交易(你必須在再次購買前出售掉之前的股票)
樣例
給出一個樣例數(shù)組 [4,4,6,1,1,4,2,5], 返回 6
solution
Approach ##1 two dp
最多可以進行兩次交易，無非就是賣出之后還要再買進一次脓杉。
所以我們用兩個dp數(shù)組分別記錄兩次交易的最大收益糟秘。
left[i]:表示從第一天到第i天進行一次交易最大的收益
right[i]:表示從第i天到最后一天進行一次交易最大的收益
填滿這兩個動態(tài)規(guī)劃數(shù)組之后，我們直接循環(huán)一次球散，找到最大值就是本題進行兩次交易的答案尿赚。
class Solution {
/**
* @param prices: Given an integer array
* @return: Maximum profit
*/
public int maxProfit(int[] prices) {
// write your code here
if (prices == null || prices.length <= 1) {
return 0;
}
int[] left = new int[prices.length];
int[] right = new int[prices.length];
// DP from left to right;
left[0] = 0;
int min = prices[0];
for (int i = 1; i < prices.length; i++) {
min = Math.min(prices[i], min);
left[i] = Math.max(left[i - 1], prices[i] - min);
}
//DP from right to left;
right[prices.length - 1] = 0;
int max = prices[prices.length - 1];
for (int i = prices.length - 2; i >= 0; i--) {
max = Math.max(prices[i], max);
right[i] = Math.max(right[i + 1], max - prices[i]);
}
int profit = 0;
for (int i = 0; i < prices.length; i++){
profit = Math.max(left[i] + right[i], profit);
}
return profit;
}
};
Approach ##2 dp
可以參考Best Time to Buy and Sell Stock IV的解析，將求出進行k次交易的最大收益蕉堰，所以只要令其k=2就是本題的解法
