121. 买卖股票最佳时期

121. 买卖股票最佳时期

• let minPrice = prices[0] 假设第一项是最小
• for (var i = 1; i < len; i++) { 遍历数组prices
• minPrice = prices[i] 重新设置最新的最小值
• max = Math.max(max, prices[i] - minPrice) 把当前值 减去最小值的差值和当前的最大值比较，选出最优

var maxProfit = function (prices) {
    var max = 0;
    var len = prices.length
    var minPrice = prices[0]
    for (var i = 1; i < len; i++) {
        if (prices[i] < minPrice) {
            minPrice = prices[i]
        } else {
            max = Math.max(max, prices[i] - minPrice)
        }
    }
    return max
};