Questionsay you had an array for which the ith element was the price of a given stock on day i.design an algorithm to find The maximum profit. Transactions at the most K. For example, given prices = [4,4,6,1,1,4,2,5], and k = 2, return 6.Answer with DP solution. LOCAL[I][J] Represents the number of 0~i, up to J transactions, and the last trade must contain PRICES[J] (that is, the last day must be sold), the maximum value of the proceeds. GLOBAL[I][J] Represents the number of 0~i, up to J transactions, the maximum value of earnings. diff = prices[i]-prices[i-1]local[i][j] = max (global[i-1][j-1] + max (diff,0), Local[i-1][j]+diff) global[i][j] = max (loc AL[I][J], global[i-1][j]) local[i-1][j] + diff means i-1 days, buy Prices[i-1], and sell for the first day. Because Local[i-1][j] indicates that the last trade must contain prices[i-1], i.e. prices[i-1] must be sold. So it is possible to combine the transaction of I-1 Day with the trading of day I, that is, the maximum of J transactions, the last day of trading must sell prices[i]. GLOBAL[I-1][J-1] indicates i-1 days, up to j-1 transactions. The last day compares diff, if diff>0, then i-1 Day buys, the first day sells, if diff<0, then I day buys, the first day sells. Class Solution {/** * @param k:an integer * @param prices:given an integer array * @return: Maximum PROFIT */public in T Maxprofit (int k, int[] prices) {//write your code here if (prices = NULL | | prices.length = 0 | | k = = 0) {RETurn 0; } int len = prices.length; if (k >= len/2) {int profit = 0; for (int i = 1; i < Len; i++) {if (Prices[i] > Prices[i-1]) {profit + = (p Rices[i]-prices[i-1]); }} return profit; } int[][] Local = new Int[len][k + 1]; int[][] Global = new Int[len][k + 1]; for (int i = 0; i < len; i++) {local[i][0] = 0; global[i][0] = 0;} Arrays.fill (Local[0], 0); Arrays.fill (Global[0], 0); for (int i = 1, i < Len; i++) {for (int j = 1; j <= K; j + +) {int diff = prices[i]-prices[i-1]; Local[i][j] = M Ath.max (Local[i-1][j] + diff, Global[i-1][j-1] + Math.max (diff, 0)); GLOBAL[I][J] = Math.max (Local[i][j], global[i-1][j]); }} return global[len-1][k]; }};
Best time to Buy and Sell Stock IV solution
