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No188BestTimeToBuyAndSellStock4.java
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No188BestTimeToBuyAndSellStock4.java
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package com.wzx.leetcode;
/**
* @see <a href="https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv/">https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv/</a>
* @author wzx
*/
public class No188BestTimeToBuyAndSellStock4 {
/**
* 股票问题动态规划统一模板
*
* dp[i][k][state]: i当前天数, k当前最多进行的交易次数, state当前是否持有股票, 值为当前收益
* dp[i][k][0] = max(dp[i-1][k][0], dp[i-1][k][1]+price[i])
* dp[i][k][1] = max(dp[i-1][k][1], dp[i-1][k-1][0]-price[i])
*/
/**
* 动态规划
* <p>
* time: O(nk)
* space: O(nk)
*/
public int maxProfit1(int k, int[] prices) {
if (k > prices.length / 2) {
// 这时相当于无限制交易次数
// 见122. best time to buy and sell stock II
int dp_i_1 = Integer.MIN_VALUE;
int dp_i_0 = 0;
for (int price : prices) {
int tmp = Math.max(dp_i_0, dp_i_1 + price);
dp_i_1 = Math.max(dp_i_1, dp_i_0 - price);
dp_i_0 = tmp;
}
return dp_i_0;
}
int[][][] dp = new int[prices.length + 1][k + 1][2];
// 边界条件, 开市前一天, 哨兵
dp[0][k][0] = 0;
for (int i = 0; i <= k; i++) {
dp[0][i][1] = Integer.MIN_VALUE;
}
for (int i = 1; i <= prices.length; i++) {
// k=0已经不允许交易, 所以没有迭代的必要
for (int j = 1; j <= k; j++) {
dp[i][j][0] = Math.max(dp[i - 1][j][0], dp[i - 1][j][1] + prices[i - 1]);
dp[i][j][1] = Math.max(dp[i - 1][j][1], dp[i - 1][j - 1][0] - prices[i - 1]);
}
}
return dp[prices.length][k][0];
}
/**
* 滚动数组
* <p>
* time: O(nk)
* space: O(k)
*/
public int maxProfit2(int k, int[] prices) {
if (k > prices.length / 2) {
// 这时相当于无限制交易次数
// 见122. best time to buy and sell stock II
int dp_i_1 = Integer.MIN_VALUE;
int dp_i_0 = 0;
for (int price : prices) {
int tmp = Math.max(dp_i_0, dp_i_1 + price);
dp_i_1 = Math.max(dp_i_1, dp_i_0 - price);
dp_i_0 = tmp;
}
return dp_i_0;
}
int[][] dp = new int[k + 1][2];
// 边界条件, 开市前一天, 哨兵
dp[k][0] = 0;
for (int i = 0; i <= k; i++) {
dp[i][1] = Integer.MIN_VALUE;
}
for (int price : prices) {
// k=0已经不允许交易, 所以没有迭代的必要
for (int j = 1; j <= k; j++) {
int tmp = Math.max(dp[j][0], dp[j][1] + price);
dp[j][1] = Math.max(dp[j][1], dp[j - 1][0] - price);
dp[j][0] = tmp;
}
}
return dp[k][0];
}
}