122. The best time to buy and sell stocks II - one-time traversal

One 、 Title Description

Give you an array of integers prices , among prices[i] Represents a stock i Sky price .

Every day , You can decide whether to buy and / Or sell shares . You at any time most Can only hold A wave of Stocks . You can also buy... First , And then in On the same day sell .

return What you can get Maximum profits .

Example 1：

 Input ：prices = [7,1,5,3,6,4]
 Output ：7
 explain ： In the  2  God （ Stock price  = 1） Buy when , In the  3  God （ Stock price  = 5） Sell when ,  The exchange will make a profit  = 5 - 1 = 4 .
      And then , In the  4  God （ Stock price  = 3） Buy when , In the  5  God （ Stock price  = 6） Sell when ,  The exchange will make a profit  = 6 - 3 = 3 .
      The total profit is  4 + 3 = 7 .

Example 2：

 Input ：prices = [1,2,3,4,5]
 Output ：4
 explain ： In the  1  God （ Stock price  = 1） Buy when , In the  5  God  （ Stock price  = 5） Sell when ,  The exchange will make a profit  = 5 - 1 = 4 .
      The total profit is  4 .

Example 2：

 Input ：prices = [7,6,4,3,1]
 Output ：0
 explain ： under these circumstances ,  There is no positive profit from trading , So you can get the maximum profit by not participating in the transaction , The biggest profit is  0 .

Two 、 Problem solving

One traverse

Find the rising range

class Solution {
    
    public int maxProfit(int[] prices) {
    
        if(prices == null){
    
            return 0;
        }
        // Use one traversal   This is a multiple transaction   And each transaction cannot be crossed 
        // So just count every rise directly   Statistical maximum profit 
        // Find the rising range 
        int length = prices.length;
        int maxProfit = 0;
        for(int i = 1;i<length;i++){
    
            if(prices[i] - prices[i-1] > 0){
    
                maxProfit += prices[i] - prices[i-1]; 
            }
        }
        return maxProfit;
    }
}