标签：fit math which return ons write [] imu let
Question 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 k transactions. For example, given prices = [4,4,6,1,1,4,2,5], and k = 2, return 6. Answer 用DP解答。 local[i][j] 表示0～i的数，最多j次交易，并且最后一次交易必须包含prices[j]（即最后一天必须卖出），收益最大值。 global[i][j] 表示0~i的数，最多j次交易，收益最大值。 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(local[i][j], global[i-1][j]) local[i-1][j] + diff 表示第i-1天，买进prices[i-1]，第i天卖出。由于local[i-1][j]表示最后一次交易必须包含prices[i-1]，即prices[i-1]必须卖出。所以可以把第i-1天的交易和第i天的交易合并，即成了最多j次交易，最后一天交易必须卖出prices[i]。 global[i-1][j-1] 表示前i-1天，最多j-1次交易。最后一天比较diff，如果diff>0，则第i-1天买进，第i天卖出；如果diff<0，则第i天买进，第i天卖出。 class Solution { /** * @param k: An integer * @param prices: Given an integer array * @return: Maximum profit */ public int 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 += (prices[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] = Math.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 解答
标签：fit math which return ons write [] imu let
原文地址：http://www.cnblogs.com/ireneyanglan/p/5988961.html
