121. Best Time to Buy and Sell Stock

题目

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: 5max. 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: 0In this case, no transaction is done, i.e. max profit = 0.

分析

找到交易价格中最大的股票利润，假设最低点为prices[i]，最高点为prices[j]，最大利润maxSum=prices[j]-prices[i]，假设i和j之间有k个中间价格，则maxSum=prices[j]-prices[x1]+prices[x1]-prices[x2]+...+prices[xk]-prices[i]，可以用动态规划解决，设向量sum记录相邻两个价格之间的最大利润，则sum[i]=max(prices[i]-prices[i-1]，sum[i-1]+prices[i]-prices[i-1])，即是按照i-1的价格出售获利大，还是按照sum[i-1]时的价格出售获利大，同时更改maxSum记录sum向量中最大的元素。

class Solution {public:    int maxProfit(vector<int>& prices) {        vector<int> sum(prices.size(),0);        int maxSum=0;        for(int i=1;i<prices.size();++i){            int temp=prices[i]-prices[i-1];            sum[i]=max(temp,sum[i-1]+temp);            maxSum=max(maxSum,sum[i]);        }        return maxSum;    }};

另一种方法是记录diff(i)为当卖出价为第i个元素时可能获得的最大利润，并且在扫描第i个数字时，能够记住前i-1个中的最小值，此时不用额外的存储空间记录获利，只需要初始化利润差，然后从第2个数字遍历，更新min值，并算出当前最大利润，与之前的最大利润之间取max即可，需要注意的是此时maxSum可能为负值，返回时判断一下。

class Solution {public:    int maxProfit(vector<int>& prices) {        if(prices.size()<2)            return 0;        int min=prices[0];        int maxSum=prices[1]-min;        for(int i=2;i<prices.size();++i){            if(prices[i-1]<min)                min=prices[i-1];            int curdiff=prices[i]-min;            maxSum=max(maxSum,curdiff);        }        return maxSum<0?0:maxSum;    }};