2014
11-19

# LeetCode-Best Time to Buy and Sell Stock III[动态规划]

### Best Time to Buy and Sell Stock III

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).

// LeetCode, Best Time to Buy and Sell Stock III
// 时间复杂度O(n)，空间复杂度O(n)
class Solution {
public:
int maxProfit(vector<int>& prices) {
if (prices.size() < 2) return 0;

const int n = prices.size();
vector<int> f(n, 0);
vector<int> g(n, 0);

for (int i = 1, valley = prices[0]; i < n; ++i) {
valley = min(valley, prices[i]);
f[i] = max(f[i - 1], prices[i] - valley);
}

for (int i = n - 2, peak = prices[n - 1]; i >= 0; --i) {
peak = max(peak, prices[i]);
g[i] = max(g[i], peak - prices[i]);
}

int max_profit = 0;
for (int i = 0; i < n; ++i)
max_profit = max(max_profit, f[i] + g[i]);

return max_profit;
}
};


Java代码:

public class Solution {
public int maxProfit(int[] prices) {
if(prices.length == 0) return 0;
int ans = 0;
int n = prices.length;

int opt[] = new int[n];
opt[0] = 0;
int low = prices[0];
int curAns = 0;
for(int i = 1; i<n; i++){
if(prices[i] < low) low = prices[i];
else if(curAns < prices[i] - low) curAns = prices[i] - low;
opt[i] = curAns;
}

int optReverse[] = new int[n];
optReverse[n - 1] = 0;
curAns = 0;
int high = prices[n - 1];
for(int i=n-2; i>=0; i--){
if(prices[i] > high) high = prices[i];
else if(curAns < high - prices[i]) curAns = high - prices[i];
optReverse[i] = curAns;
}

for(int i=0; i<n; i++){
int tmp = opt[i] + optReverse[i];
if(ans < tmp) ans = tmp;
}
return ans;
}

}