returnmaxP; 对于iii: 同样不需要额外的空间,而且只需要O(n)的空间复杂度。思路就是我一开始没有钱,buy1是我第一次买股票之后剩的钱(肯定是负的,我白手起家,借钱买股票啊),sell1是我卖完第一次买的股票之后剩的钱(>=0),buy2是我用上回挣得钱(有可能需要找别人再借一点)买第二次股票,sell2是我卖完第...
publicintmaxProfit(int[]prices){if(prices.length==0){return0;}intK=2;int[][]dp=newint[prices.length][K+1];for(intk=1;k<=K;k++){intmin=prices[0];for(inti=1;i<prices.length;i++){//找出第 1 天到第 i 天 prices[buy] - dp[buy][k - 1] 的最小值min=Math.min(prices[i]...
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...
Best Time to Buy and Sell Stock III Question Solution 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...
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 (i.e., you must sell the stock bef...
Can you solve this real interview question? Best Time to Buy and Sell Stock III - You are given an array prices where prices[i] is the price of a given stock on the ith day. Find the maximum profit you can achieve. You may complete at most two transacti
Can you solve this real interview question? Best Time to Buy and Sell Stock III - You are given an array prices where prices[i] is the price of a given stock on the ith day. Find the maximum profit you can achieve. You may complete at most two transacti
123. Best Time to Buy and Sell Stock III,理一下思路:二维数组l表示到第i天为止最多进行j次交易并且最后一次交易在最后一次交易在最后一天卖出的最大利润g表示到第i天为止最多进行j次交易的最大利润第i天卖第j支股票...
当k = 无数次时,即这一题@&再见萤火虫&【LeetCode_Array_122. Best Time to Buy and Sell Stock II买股票的最佳时机(C++)】。 解决这题时有点类似于贪心算法,同样是记录卖出前的最低价格minPrice,但是一旦遇到 prices[i] 卖出时,利润少于上一笔 prices[i-1] 的利润,就果断按照上一笔的...
Best Time to Buy and Sell Stock III 高级版的难度一下子就提升了,刚一开始我的错误想法是:将两个获益最大区间的获益相加,后来很快就证明了这个解法是错误的。 借助网络之后,确定了正确解法:构造两个数组,left和right,left[i]表示从0到i天的最大获益,right[i]表示从i到最后一天的最大获益。