letecode [121] - Best Time to Buy and Sell Stock

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 (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Note that you cannot sell a stock before you buy one.

Example 1:

Input: [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 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
Explanation: In this case, no transaction is done, i.e. max profit = 0.

 

题目大意：

　　给定一组值表示某股票这些天的价格，在某天买入，在之后的某天卖出，求利润最大值。

理解：

　　方法一：暴力法。求从1到n-1买入，2到n天卖出的利润最大值。非常耗时，复杂度O(n*n)。

　　方法二：如该图，用min记录当前最小的买入值（保证遍历时该卖出值对应它的最大利润），dif记录当前最大利润，遍历数组，更新min与dif。

　　　　　　注意：若卖出值<买入值，更新买入值，且不影响dif。

代码C++:方法一：

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

方法二：

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

运行结果：

　　方法一：执行用时 : 1656 ms　　内存消耗 : 9.6 MB

　　方法二：执行用时 : 8ms　　内存消耗 : 9.6 MB