LeetCode 买卖股票的最佳时机II

题目

示例

思路

买入的时候从最前面开始寻找最小的元素，以最低价格买入，卖出的时候必须在持有股票，并且寻找后面的极大的元素，然后在这个时候卖出。

代码

int maxProfit(int* prices, int pricesSize)
{    
    int buy = 0 , sell = 0, sum=0, i=0;    
    while (i < pricesSize)     
    {        
       if (buy == 0)         
       {               
          if (i+1 < pricesSize && prices[i] < prices[i+1])             
          {                
             buy = prices[i];                
             sell = prices[i+1];            
          }        
       }         
       else         
       {             
          if (i+1 >= pricesSize ||prices[i] > prices[i+1])           
          {                 
             sum += sell-buy;                
             buy = 0;            
          }             
          else  
             sell = prices[i+1];         
       }        
       i++;    
    }    
    return sum;
}

这里股票的价格是有可能取0的，因此在代码中这条语句是会漏掉交易的**if (buy == 0)**,但股票价格是不会为负的，因此在没有持有股票的时候，对 buy 赋值为-1。

改善

int maxProfit(int* prices, int pricesSize)
{    
    int buy = -1 , sell = 0, sum=0, i=0;    
    while (i < pricesSize)     
    {        
       if (buy == -1)         
       {               
          if (i+1 < pricesSize && prices[i] < prices[i+1])             
          {               
             buy = prices[i];                
             sell = prices[i+1];            
          }        
       }         
       else         
       {            
          if (i+1 >= pricesSize ||prices[i] > prices[i+1])             
          {                 
             sum += sell-buy;                
             buy = -1;            
          }             
          else                
             sell = prices[i+1];         
       }        
       i++;    
    }    
    return sum;
}