First, the best time to buy and sell stocks array ---

Given an array, which i-th element is a given stock i-day prices.

If you are only allowed up to complete a transaction (ie buying and selling a stock), to design an algorithm to compute the maximum profit you can get.

Note that you can not sell the stock before buying stocks.

Example 1:

Input: [7,1,5,3,6,4]
Output: 5
Explanation: Day 2 (= 1 stock price), when buying, on day 5 (stock price = 6), when sold, The maximum profit = 6-1 = 5.
Note that profits can not be 7-1 = 6, because the selling price must be greater than the purchase price.
Example 2:

Enter: [7,6,4,3,1]
Output: 0
to explain: In this case, no transaction is completed, the maximum profit is zero.

 

Method a: two traverse

Time and space consumption! !

 1 class Solution {
 2 public:
 3     int maxProfit(vector<int>& prices) {
 4         int max_value = 0;
 5         int res;
 6         if(prices.size()<=1) return 0;
 7         for(int i=prices.size() - 1;i>=0;i--){
 8             for(int j=i-1;j>=0;j--){
 9                 res = prices[i] - prices[j];
10                 if(res > max_value) max_value = res; 
11             }
12         }
13         return max_value;
14     }
15 };

 

Method two: into sequences and a maximum seek

 1 class Solution {
 2 public:
 3     int maxProfit(vector<int>& prices) {
 4         vector<int> trend;
 5         if (prices.size() < 2)
 6         {
 7             return 0;
 8         }
 9         for (int i = 0; i < prices.size() - 1; i++)
10         {
11             trend.push_back(prices[i+1] - Prices [I]);   // The daily stock price ups and downs of the stock configuration array 
12 is          }
 13 is          
14          int SUM = 0 ;
 15          int max = Trend [ 0 ];
 16          for ( int I = 0 ; I <Trend. size (); I ++)          // maximum Change subarray arrays and calculating 
. 17          {
 18 is              SUM = SUM> 0 SUM +? Trend [I]: Trend [I];
 . 19              ? max = max> SUM max: SUM;
 20 is          }
 21 is          max = max < 0 ? 0 : max;
22         return max;
23     }
24 };