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.分析
首先通过把相邻的股票价格相减得到一组差价，后面就转化成了最大子列和问题了；另外注意当传进来的prices的数组为空时，返回0；
class Solution {
public:
int maxProfit(vector<int>& prices) {
if(prices.size()==0) return 0;
vector<int> profits(prices.size());
profits[0]=0;
for(int i=1;i<profits.size();i++)
profits[i]=prices[i]-prices[i-1];
int maxsum[profits.size()]={0},maxprofit=0;
for(int i=1;i<profits.size();i++){
maxsum[i]=max(profits[i],maxsum[i-1]+profits[i]);
maxprofit=max(maxprofit,maxsum[i]);
}
return maxprofit;
}
};