LeetCode —— Best Time to Buy and Sell Stock III

链接：http://leetcode.com/onlinejudge#question_123

原题：

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).
思路：简单的动态规划，因为限定最多只能有两次交易，而且交易不能重叠。那么分两步走好了，

由左向右记录一下到那一天为止最大的收益；同样，由右到左，记录最大收益。

然后加一加就可以了。

代码：

class Solution {
public:
int maxProfit(vector<int> &prices) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int size = prices.size();
if (size <= 1)
return 0;

vector<int> leftIncomes(size, 0);
vector<int> rightIncomes(size, 0);

int min = prices[0];
for (int i=1; i<size; i++) {
if (prices[i] < min)
min = prices[i];
if (prices[i] - min > leftIncomes[i-1])
leftIncomes[i] = prices[i] - min;
else
leftIncomes[i] = leftIncomes[i-1];
}

int max = prices.back();
for (int i=size-2; i>=0; i--) {
if (prices[i] > max)
max = prices[i];
if (max - prices[i] > rightIncomes[i+1])
rightIncomes[i] = max - prices[i];
else
rightIncomes[i] = rightIncomes[i+1];
}

int profit = 0;
for (int i=0; i<size; i++) {
if (leftIncomes[i] + rightIncomes[i] > profit)
profit = leftIncomes[i] + rightIncomes[i];
}

return profit;
}
};