Question ：    

 

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.

for example： array[]  = { 2, 5, 3, 8, 9, 4 } ， maxProfit = 9 - 2 = 7.

Anwser 1 ：       

 

class Solution {public:    int maxProfit(vector
    
      &prices) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if(prices.size() == 0) return 0;                int ret = 0;                int len = prices.size();        int maxPrice = prices[len-1];        for(int i = len - 1; i >= 0; i--){            maxPrice = max(prices[i], maxPrice);    // maxPrice            ret = max(ret, maxPrice - prices[i]);   // maxProfit        }                return ret;    }};
    

注意点：

 

最大利润，应该是先买的最低价与后卖的最高价的差值，因此需要考虑时间先后顺序

 

Anwser 2 ：       

 

class Solution {public:    int maxProfit(vector
    
      &prices) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        int maxp = 0;        int dp = 0;                for(int i = prices.size()-2; i >= 0 ;i--)        {            if(dp >= 0){                dp += (prices[i+1] - prices[i]);            } else {                dp = max(0, prices[i+1] - prices[i]);            }            maxp = max(dp, maxp);        }          return maxp;    }};
    

说明：

 

1） 此法把两数之间最大差，转化为了求两数组之间最大和

2） dp += (prices[i+1] - prices[i]) 实际上是 dp +=  (prices[i+1] - prices[i])  +  (prices[i] - prices[i-1])  +  (prices[i-1] - prices[i-2]) + ... =  (prices[i] - prices[j])   （i > j）

 

 

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