算法作业_40(2017.6.22第十八周)

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: 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.

题意：用一个数组表示股票每天的价格，数组的第i个数表示股票在第i天的价格。 如果只允许进行一次交易，也就是说只允许买一支股票并卖掉，求最大的收益。

分析：动态规划法。从前向后遍历数组，记录当前出现过的最低价格，作为买入价格，并计算以当天价格出售的收益，作为可能的最大收益，整个遍历过程中，出现过的最大收益就是所求。

class Solution {public:    int maxProfit(vector<int>& prices) {        int min  = 0 ;        int res  = 0 ;        int n = prices.size();        if(n==0 || n==1) return 0;                vector<int> dp (n);        min = prices[0];        dp[0] = 0;                for(int i = 1 ; i <n ; i++){            if(prices[i]<min){                min = prices[i];                dp[i] = 0;            }else{                dp[i] = prices[i] -min;            }        }        for(int i = 0 ;i <n;i++){            if(res<dp[i]) res = dp[i];        }        return res ;    }};