Best Time to Buy and Sell Stock

标签：leetcode   动态规划   股票收益最大   java   

问题一 Best Time to Buy and Sell Stock

public int maxProfit(int[] prices) {
		assert prices.length >= 0;
		int min = Integer.MAX_VALUE;
		int max = 0;// 寻找最小的，一次保存最大的
		for (int p : prices) {
			min = Math.min(min, p);
			max = Math.max(max, p - min);
		}
		return max;
	}

public int maxProfit(int[] prices) {
		assert prices.length >= 0;
		int local = 0;
		int max = 0;
		for (int i = 1; i < prices.length; i++) {
			local = prices[i] - prices[i - 1];
			int x = local > 0 ? local : 0;
			max += x;
		}
		return max;
	}

public int maxProfit(int[] prices) {
		if (prices.length < 2)
			return 0;
		int[] left = new int[prices.length];// 记录了prices[0..i]的最大profit
		int[] right = new int[prices.length];// 记录了prices[i..n]的最大profit
		int left_min = prices[0];
		for (int i = 1; i < prices.length; i++) {
			left_min = Math.min(left_min, prices[i]);
			left[i] = Math.max(prices[i] - left_min, left[i - 1]);
		}
		int right_max = prices[prices.length - 1];
		for (int i = prices.length - 2; i >= 0; i--) {
			right_max = Math.max(right_max, prices[i]);
			right[i] = Math.max(right_max - prices[i], right[i + 1]);
		}
		int max = 0;
		for (int i = 1; i < prices.length; i++) {
			max = Math.max(max, left[i] + right[i]);
		}
		return max;
	}

这时可以定义一个2位数组，用来保存买入和卖出的index，当全局的收益有变化时，更新即可。

public int[] maxProfits(int[] prices) {
		assert prices.length >= 0;
		int min = Integer.MAX_VALUE;
		int[] res = new int[2];// 最终保存的最大
		int[] temp = new int[2];// index的临时的记录
		int max = 0;
		for (int i = 0; i < prices.length; i++) {
			if (min > prices[i]) {
				min = prices[i];
				temp[0] = i;
			}
			if (max < prices[i] - min) {
				temp[1] = i;
				max = prices[i] - min;
				res = Arrays.copyOf(temp, 2);
			}
		}
		System.out.println(res[0] + "----" + res[1]);
		return res;
	}

Best Time to Buy and Sell Stock

标签：leetcode   动态规划   股票收益最大   java   

原文地址：http://blog.csdn.net/lgcssx/article/details/41629439