188. Best Time to Buy and Sell Stock IV
Say you have an array for which the _i_th 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.
Note: You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
Example 1:
Input: [2,4,1], k = 2 Output: 2 Explanation: Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit = 4-2 = 2.
Example 2:
Input: [3,2,6,5,0,3], k = 2 Output: 7 Explanation: Buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4. Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
思路：
如果会做123题，这一题就一定能想到这个最优解，也是动态规划，题目是允许交易k次，而123题是只允许两次。
代码：
java：
代码语言：javascript复制class Solution {
public int maxProfit(int k, int[] prices) {
int len = prices.length;
if (len == 0) return 0;
if (k >= len/2) return helper(prices);
int[] dp = new int[k 1];
int[] min = new int[k 1];
for (int i = 0; i <min.length; i ) {
min[i] = prices[0];
}
for (int i = 0; i < prices.length; i ) {
for (int j = 1; j <= k; j ) {
dp[j] = Math.max(dp[j], prices[i] - min[j]);
min[j] = Math.min(min[j], prices[i] - dp[j-1]);
}
}
return dp[k];
}
private int helper(int[] p) {
int len = p.length;
int res = 0;
for(int i = 1; i < len; i ) {
if (p[i] > p[i-1]) res = (p[i] - p[i-1]);
}
return res;
}
}
