LeetCode - Best Time to Buy and Sell Stock III

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

Design an algorithm to find the maximum profit. You may complete at most two transactions.

Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

Solution 1:

public int maxProfit(int[] prices) {
    int n = prices.length;
    if(n<=1) return 0;
    
    // dp数组保存左边和右边的利润最大值
    int[] leftProfit = new int[n]; // [0,i]区间的最大利润
    int[] rightProfit = new int[n]; // [i,n-1]区间的最大利润
    
    //正向遍历，leftProfit[i]表示 prices[0...i]内做一次交易的最大收益.
    int minPrice = prices[0]; //最低买入价
    for(int i=1; i<n; i++) {
        minPrice = Math.min(minPrice, prices[i]); // 更新最小买入价  
        // i的最大利润为（i-1的利润）和（当前卖出价和之前买入价之差）的较大者
        leftProfit[i] = Math.max(leftProfit[i-1], prices[i]-minPrice);
    }
    
    //逆向遍历, rightProfit[i]表示 prices[i...n-1]内做一次交易的最大收益.
    int maxPrice = prices[n-1]; //最高卖出价
    for(int i=n-2; i>=0; i--) {
        maxPrice = Math.max(maxPrice, prices[i]); // 更新最高卖出价  
        // i的最大利润为（i+1的利润）和（最高卖出价和当前买入价之差）的较大者
        rightProfit[i] = Math.max(rightProfit[i+1], maxPrice-prices[i]);
    }
    
    int maxProfit = 0;
    for(int i=0; i<n; i++) {
        maxProfit = Math.max(maxProfit, leftProfit[i]+rightProfit[i]);
    }
    return maxProfit;
}

 

Solution 2:

忘记上面的方法吧，下面这个解法更好。

public int maxProfit(int[] prices) {
    int len = prices.length;
    if (len == 0) return 0;
	int k = 3; // 2 transactions
	int[][] f = new int[k][len];
	for (int i = 1; i < k; i++) {
		int tmpMax = f[i-1][0] - prices[0];
		for (int j = 1; j < len; j++) {
			f[i][j] = Math.max(f[i][j-1], tmpMax + prices[j]);
			tmpMax = Math.max(tmpMax, f[i-1][j] - prices[j]);
		}
	}
	return f[k-1][len-1];
}