Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
[balabala] 此题参照 网友博客的解法 http://www.cnblogs.com/grandyang/p/4295761.html。需要维护两个dp数组,分别是local[i][j] 和global[i][j],前者表示到第i天为止至多完成j次交易且最后一笔交易于第i天完成的最大收益,是一个局部最优值,后者表示到第i天为止至多完成j次交易获取的最大值, 为全局最优值。global[i][j] = max(global[i - 1][j], local[i][j]), 这个递推式比较好理解,是按照第 i 天是否参与交易分类讨论的,local[i][j] = max(global[i - 1][j - 1] + max(0, diff), local[i - 1][j] + diff).
对于这个解法一度半懂不懂,为什么需要两个dp数组呢?local[][]数组的递推公式如何理解? 在理解的过程中发现可以这样去理解那些不理解的事物(说得有些绕):去掉看看效果,正确的解法里每个变量都是有其不可替换的作用的。假设我们只用一个dp数组,dp[i][j]含义同global[i][j],表示到第i天至多完成j次交易能获取的最大利润,自然地,可以按照第 i 天是否参与最后一笔交易分类讨论,于是递推公式dp[i][j] = max(dp[i - 1][j], max(dp[k][j - 1] + maxProfitBetween(k + 1, i)) ), 式中第二项表示 第i天参与交易的情况,看到仅有dp[][]数组的情况下写起来是多么复杂。于是理解了local[][]数组的好处,使算法复杂度由O(n^3)变为O(n^2)( 对比Palindrome Partition II 的降维思路,目前还没看出两种降维思路的想通之处)。local[i][j]是如何递推的呢? 它的作用是记录到第 i 天至多完成 j 次交易的最大利润且第 i 天完成最后一笔交易,这最后一笔交易卖出点是day i, 买入点可以是0 - i的任意一天,分三种情况讨论:
1) 在i - 1天之前买入:这种情况下可将最后一笔交易以day[i-1]为分割点分解为两笔交易来计算,第一笔是在day[k](k < i -1)买入,day[i -1]卖出,对应于local[i - 1][j], 第二笔是day[i - 1]买入,day[i]卖出,火力为prices[i] - prices[i - 1], 于是这种情况下local[i][j] = local[i-1][j] + prices[i] - prices[i - 1]
2) 在第 i - 1天买入:local[i][j]= global[i - 1][j - 1] + prices[i] - prices[i - 1]
3) 在第 i 天买入: local[i][j]= global[i - 1][j - 1] + 0
合并起来,local[i][j] = local[i][j] = max(global[i - 1][j - 1] + max(0, diff), local[i - 1][j] + diff), 其中diff = prices[i] - prices[i - 1]
第三个实现yb君的思路,另一角度看问题~
public class Solution { // Method 3: space: O(N * k), time: O(N * k) public int maxProfit(int k, int[] prices) { if (prices == null || prices.length == 0) return 0; int N = prices.length; if (N <= k - 1) return maxProfitWithoutLimit(prices); int M = 2 * k; // dp[i][j]: on day i (1, 2... N), max profit after transaction j done // transaction j means buy j / 2 + j % 2 times, sell j / 2 times int[][] dp = new int[N + 1][M + 1]; for (int i = 1; i <= N; i++) { for (int j = 1; j <= i && j <= M; j++) { if (j % 2 == 1) {// buy dp[i][j] = dp[i - 1][j - 1] - prices[i - 1]; } else { // sell dp[i][j] = dp[i - 1][j - 1] + prices[i - 1]; } if (i - 1 >= j) dp[i][j] = Math.max(dp[i][j], dp[i - 1][j]); } } int max = 0; for (int j = 2; j <= M; j += 2) { max = Math.max(max, dp[N][j]); } return max; } // Method 2space: O(k), time: O(N * k) public int maxProfit(int k, int[] prices) { if (prices == null || prices.length == 0) return 0; int days = prices.length; if (k >= days) return maxProfit(prices); // local[j] 表示到当日至多完成j次交易,并且最后一笔交易在当日完成的最大收益 int[] local = new int[k + 1]; // global[j] 表示到当日至多完成j次交易的最大收益 int[] global = new int[k + 1]; for (int i = 1; i < days; i++) { for (int j = k; j > 0; j--) { // 逆序以利用前一天的结果 int diff = prices[i] - prices[i - 1]; local[j] = Math.max(global[j - 1] + Math.max(diff, 0), local[j] + diff); global[j] = Math.max(global[j], local[j]); } } return global[k]; } // Method 1: space: O(N * k), time: O(N * k) public int maxProfit1(int k, int[] prices) { if (prices == null || prices.length == 0) return 0; int days = prices.length; if (k >= days) return maxProfit(prices); // local[i][j] 表示到第i天至多完成j次交易,并且最后一笔交易在第i天完成的最大收益 int[][] local = new int[days][k + 1]; // global[i][j] 表示到第i天至多完成j次交易的最大收益 int[][] global = new int[days][k + 1]; for (int i = 1; i < days; i++) { local[i][0] = 0; global[i][0] = 0; for (int j = 1; j <= k; j++) { int diff = prices[i] - prices[i - 1]; local[i][j] = Math.max(global[i - 1][j - 1] + Math.max(diff, 0), local[i - 1][j] + diff); global[i][j] = Math.max(global[i - 1][j], local[i][j]); } } return global[days - 1][k]; } private int maxProfit(int[] prices) { if (prices == null || prices.length == 0) return 0; int days = prices.length; int profit = 0; int delta = 0; for (int i = 1; i < days; i++) { delta = prices[i] - prices[i - 1]; profit += delta > 0 ? delta : 0; } return profit; } }