JS 力扣刷题 121. 买卖股票的最佳时机

var maxProfit = function(prices) {
    //保留之前最小值
    let min = Infinity;
    //算当前值与最小值的差值
    let res = 0;
    for(let i = 0; i < prices.length; i++){
        if(min > prices[i])min = prices[i];
        if(res < prices[i] - min)res = prices[i] - min;
    }
    return res;
};