leetcode——121.买卖股票的最佳时机

 

 

用时8分20秒：

public int maxProfit(int[] prices) {
        int n = prices.length;
        if(n<=1) return 0;
        int min_price = prices[0];
        int interest = 0;
        for(int i = 1;i<n;i++){
            min_price = Math.min(min_price,prices[i]);
            interest = Math.max(interest,prices[i] - min_price);
        }
        return interest;
    }

 做了如下修改，性能得到很大提升：

public int maxProfit(int[] prices) {
        int n = prices.length;
        if(n<=1) return 0;
        int min_price = prices[0];
        int interest = 0;
        for(int i = 1;i<n;i++){
            if(prices[i] < min_price) {
                min_price = prices[i];
            }else{
                interest = Math.max(interest,prices[i] - min_price);
            }
        }
        return interest;
    }

 ——2020.8.5