121. Best Time to Buy and Sell Stock

Say you have an array for which the i^th 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.

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最大的股票收益问题

买进卖出，求最大收益

贪心法，分别找到价格最低和最高的一天，低进高出，

利用一个变量profit记录，目前最大收益

利用一个变量cur_min记录，目前为止最小的价格

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也可以把原始价格序列变成差分序列，本题也可以最大m字段和，m = 1

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code如下：

class Solution {
public:
    //只能买卖一次股票，取得最大值，应该在最低点买，在最高点卖
    int maxProfit(vector<int>& prices) {
        int length = prices.size();
        int maxProfit = 0;//到目前为止，能得到的最大的利润
        int minprice;//到目前位置，最低的价格
        for(int i = 0;i<length;i++){
            maxProfit = max(maxProfit,prices[i]-minprice);
            minprice = min(minprice,prices[i]);
        }
        return maxProfit;
    }
};

 

转载于:https://www.cnblogs.com/li-daphne/p/5625122.html