Best Time to Buy and Sell Stock

标签：best time to buy and   leetcode   

Say you have an array for which the i^th element is the price of a given stock on day i.

If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

问题抽象出来就是找到数组最大差值，并且满足小在前大在后条件。

首先不加思考的写了下面这段代码测试下Leetcode是否有严格时间要求:

public class Solution {
        public int maxProfit(int[] prices) {
        if(prices.length==0){
        	return 0;
        }
        int MaxProfit = 0;
        int len = prices.length;
        for(int i=0;i<len;i++){
        	for(int j=i+1;j<len;j++){
        		if(prices[j]-prices[i]>MaxProfit){
        			MaxProfit = prices[j]-prices[i];
        		}
        	}
        }
        return MaxProfit;
    }
}

不出意外TLE。

其实稍微思考就会发现两个循环是很大的时间浪费，其实每次我们只需要找到当前位置最小值，往后如果比最小值小更新最小值，否则进行最大差值计算。

public class Solution {
       public int maxProfit(int[] prices) {
        if(prices.length==0){
        	return 0;
        }
        int MaxProfit = 0;
        int len = prices.length;
        int Min = prices[0];
        for(int i=1;i<len;i++){
        	if(prices[i]<Min){
        		Min = prices[i];
        	}
        	else {
				if(prices[i]-Min>MaxProfit){
					MaxProfit = prices[i]-Min;
				}
			}
        }
        return MaxProfit;
    }
}

Best Time to Buy and Sell Stock

标签：best time to buy and   leetcode   

原文地址：http://blog.csdn.net/huruzun/article/details/39429631