leetcode top 100 的121_BestTimetoBuyandSellStock

这是leetcode top 100 的121_BestTimetoBuyandSellStock。

Say you have an array for which the ith 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.

Example 1:

Input: [7, 1, 5, 3, 6, 4]
Output: 5

max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)

Example 2:

Input: [7, 6, 4, 3, 1]
Output: 0

In this case, no transaction is done, i.e. max profit = 0.

package leetcodeJava.array;

public class top100_121_BestTimetoBuyandSellStock {
	
//	public static int maxProfit(int[] nums){
//		int i=0;
//		int j=nums.length-1;
//		int buy=nums[0];
//		int sell=-1;
//		while (i<j)
//		{
//			i++;
//			if (nums[i]<buy){
//				buy=nums[i];
//			}
//			
//			if (nums[j]>sell){
//				sell=nums[j];
//				
//			}
//		}
//		
//		
//		return 0;
//	}
	
	public static int maxProfit(int[] prices){
		
		if (prices==null || prices.length<1)
			return 0;
		
		int minBuy=prices[0];	//用minBuy记录最小买入价格
		int maxProfit=0;		//用maxProfit记录最大的利润
		
		for(int i=1;i<prices.length;i++){
			//如果当天的价格更适合当做买入价格的话，则赋给minBuy
			if (minBuy>prices[i]){
				minBuy=prices[i];
			}
			//如果不适合的话，就看看利润是否有变化
			else {
				if (maxProfit<(prices[i]-minBuy))
					maxProfit=prices[i]-minBuy;
			}
		}
		
		return maxProfit;
	}
	
	public static void main(String [] args){
		int [] prices={7,1,5,3,0,4};
		System.out.println(maxProfit(prices));
	}

}