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. 求两数之差的最大值，要求小的数在大的数前面 C++（9ms）：

1 class Solution { 2 public: 3     int maxProfit(vector
     
      & prices) { 4         int len = prices.size() ; 5         if (len < 1) 6             return 0 ; 7         int minPri = prices[0] ; 8         int maxDiff = 0  ; 9         for (int i = 0 ; i < len ; i++){10             minPri = min(minPri , prices[i]) ;11             maxDiff=  max(maxDiff , prices[i] - minPri) ;12         }13         return maxDiff ;14     }15 };
     

 

java(3ms):

1 class Solution { 2     public int maxProfit(int[] prices) { 3         int len = prices.length ; 4         if (len < 1) 5             return 0 ; 6         int Min = prices[0] ; 7         int Max = 0 ; 8         for(int i = 0 ; i < len ; i++){ 9             Min = Math.min(Min,prices[i]) ;10             Max = Math.max(Max, prices[i]-Min) ;11         }12         return Max ;13     }14 }