题解 | #买卖股票的最好时机#

let max = 0
let a =0
let b 

for(let i =2 ;i<=prices.length;i++) {

    if(a >=0 && prices[i-1] - prices[i-2] >=0) {
        b =a + prices[i-1] - prices[i-2]
    }else{
        b = Math.max(a+ prices[i-1] - prices[i-2],0)
    }

    max = Math.max(b,max)
    a = b
}

return max