[LeetCode] Best Time to Buy and Sell Stock

Say you have an array for which the ith element is the price of a given stock on day i.

If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Example 1:

Input: [7, 1, 5, 3, 6, 4]
Output: 5

max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price) 

Example 2:

Input: [7, 6, 4, 3, 1]
Output: 0

In this case, no transaction is done, i.e. max profit = 0.

只进行一次交易，使得到最大的利润。首先想到使用蛮力算法两层for循环遍历数组。结果导致超时。

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if (prices.size() == 0 || prices.size() == 1)
            return 0;
        int res = 0, mindiff = INT_MAX;
        for (int i = 0; i != prices.size() - 1; i++) {
            for (int j = i + 1; j != prices.size(); j++) {
                int tmp = prices[j] - prices[i];
                if (tmp >= 0)
                    res = max(res, tmp);
            }
        }
        return res;
    }
};

接着想了一个办法就是遍历一次数组。在遍历的同时比较出最小的购买价格，然后用这一天以后的每天价格减去这个最小购买价格取最大值即可。

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if (prices.size() == 0 || prices.size() == 1)
            return 0;
        int res = 0, minDiff = INT_MAX;
        for (int i = 0; i != prices.size(); i++) {
            minDiff = min(minDiff, prices[i]);
            res = max(res, prices[i] - minDiff);
        }
        return res;
    }
};
// 6 ms

使用DP也可以求解这个问题，用dp[i]数组存储截止第i天的最大利润值。状态转移方程是 dp[i] = max(dp[i - 1], prices[i] - minPrice)

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if (prices.size() == 0)
            return 0;
        int n = prices.size();
        vector<int> dp(n, 0);
        int minPrice = prices[0];
        for (int i = 0; i != n; i++) {
            dp[i] = max(dp[i - 1], prices[i] - minPrice);
            minPrice = min(minPrice, prices[i]);
        }
        return dp[n - 1];
    }
};
// 6 ms

相关题目：Best Time to Buy and Sell Stock II