Best Time to Buy and Sell Stock III - LeetCode 123

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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Array Dynamic
Programming

Hide Similar Problems
(M) Best Time to Buy and Sell Stock (M)
Best Time to Buy and Sell Stock II (H) Best Time to Buy and Sell Stock IV

分析：

最多买卖两次，那就把元素组分成两段，分别求出两段的最大利益，然后相加，于是得到动态规划解法（类似于求出数组中除去自身元素外的其他所有元素的乘积）：

先从前往后，求出第0天到第i天买卖一次能获得的最大利益，记入forward[i];

再从后往前，求出第i天到n-1 天买卖一次能获得的最大利益，记入backward[i];

然后两次买卖的最大利益 max_profi = max{forward[i] + backward[i]},其中i取值[0,n-1]

////////////////12ms//////////////////////////////////
class Solution {
public:
/**
* @param prices: Given an integer array
* @return: Maximum profit
*/
int maxProfit(vector<int> &prices) {
// write your code here
int n = prices.size();
if(n < 2)
return 0;
if( n == 2)
return prices[1]> prices[0]? prices[1] - prices[0]:0;
vector<int> forward(n,0); //forward[i]表示从第0天到第i天买卖一次能获得的最大利益
vector<int> backward(n,0); //backward[i]表示从第i天到n-1天买卖一次能获得的最大利益

int min_p = prices[0];
for(int i = 1; i < n; i++){  //从前往后计算forward 
if(prices[i] < min_p)
min_p = prices[i];
int t = prices[i] - min_p;
forward[i] =  max(t,forward[i-1]);
}

int max_p = prices[n-1];
for(int i = n-2; i >= 0; i--){  //从后往前计算backward
if(prices[i] > max_p)
max_p = prices[i];
int t =  max_p - prices[i];
backward[i] = max(t,backward[i+1]);
}

int max_profi = 0;
for(int i = 0; i < n - 1; i++){ //求出两段和最大值
int t = forward[i] + backward[i];
max_profi = max(max_profi , t);
}

return max_profi;
}
};