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LeetCode--Best Time to Buy and Sell Stock (贪心策略 or 动态规划)

Best Time to Buy and Sell Stock

 
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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.

Have you been asked this question in an interview? 



解题思路:

贪心法:

分别找到价格最低和最高的一天,低进高出,注意最低的一天要在最高的一天之前。

把原始价格序列变成差分序列,也就是对于元素j,在该点卖出的价格等于prices[j] 减去它前面元素中最小的值。这样遍历一遍数组就OK了。

本题也可以做是最大m 子段和,m = 1 。


动态规划法:

首先用prices[i] - prices[i-1] 将数组转化为 差分序列,那么问题就转化为在这个差分序列中找出一个最大( maximum )的连续的子序列和的问题,可以用http://blog.csdn.net/acema/article/details/26979779 类似方法求解。


备注:若采用暴利破解法,时间复杂度是O(n^2),会超出时间限制。




程序源代码:

贪心算法Java代码

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


动态规划:java代码

    public int maxProfit(int[] prices) {
        int profit=0;
        int temp=0;
        for(int i=1; i<prices.length; ++i){
        	int diff=prices[i] - prices[i-1];
        	temp=Math.max(temp+diff, diff);
        	profit=Math.max(profit, temp);
        }
        return profit;
    }



暴利破解法:

	
	/**
	 * 暴利枚举法,超出时间限制
	 * @param prices
	 * @return
	 */
	public static  int maxProfit2(int[] prices) {
        int profit= 0;
        for( int i=0; i<prices.length-1; ++i){
        	for (int j=i+1; j<prices.length; ++j){
        		profit = Math.max(prices[j]-prices[i], profit);
        	}
        }
        return profit;
    }