动态规划、背包问题 6/23 101-105

1.买卖股票的最佳时机（ LeetCode 121 ）

动态规划

   class Solution {
    
        public int maxProfit(int[] prices) {
    
            if(prices.length <= 1)return 0;
            int max = 0;//最大利润
            int min = prices[0];//min维护当天之前的最下的一天的股票价格
            for (int i = 1; i < prices.length; i++) {
    
                max = Math.max(max,prices[i] - min);
                if (prices[i] < min)min = prices[i];
            }
            return max;
        }
    }

2.买卖股票的最佳时机II（ LeetCode 122 ）

方法一：贪心算法

  class Solution {
    
        public int maxProfit(int[] prices) {
    
            if (prices.length <= 1)return 0;
            boolean flag = false;//是否持有股票
            int ans = 0;
            int pre = 0;//手中刚买的股票
            for (int i = 0; i < prices.length; i++) {
    
                if (!flag){
    //当前手中未持股
                    flag = true;
                    pre = prices[i];//买入股票
                }else if (prices[i] < pre){
    //今天更便宜今天买入
                    pre = prices[i];
                }else if (i + 1 < prices.length && prices[i + 1] < prices[i]){
    //明天卖出会更少赚一点
                    ans += prices[i] - pre;
                    flag = false;
                }else if (i == prices.length - 1){
    
                    ans += prices[i] - pre;
                    flag = false;
                }
            }

            return ans;
        }
    }

方法二：官方题解，贪心算法

class Solution {
    
    public int maxProfit(int[] prices) {
    
        int ans = 0;
        int n = prices.length;
        for (int i = 1; i < n; ++i) {
    
            ans += Math.max(0, prices[i] - prices[i - 1]);
        }
        return ans;
    }
}

方法三：动态规划
维护两个状态

class Solution {
    
        public int maxProfit(int[] prices) {
    
            if (prices.length <= 1) return 0;
            int[][] dp = new int[prices.length][2];
            dp[0][0] = 0;//第0天结束未持有股票的最大收益
            dp[0][1] = -prices[0];//第0天结束持有股票的最大收益
            for (int i = 1; i < prices.length; i++) {
    
                dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][1] + prices[i]);
                dp[i][1] = Math.max(dp[i - 1][1], dp[i - 1][0] - prices[i]);
            }
            //最后一天未持有股票肯定更大
            return dp[prices.length - 1][0];
        }
    }

方法四：动态规划 优化

 class Solution {
    
        public int maxProfit(int[] prices) {
    
            if (prices.length <= 1) return 0;
           int x = 0;//当天未持有股票最大收益
           int y = - prices[0];//当天持有股票最大收益
            for (int i = 1; i < prices.length; i++) {
    
                x = Math.max(x, y + prices[i]);
                y = Math.max(y, x - prices[i]);
            }
            //最后一天未持有股票肯定更大
            return x;
        }
    }

3.买卖股票的最佳时机III（ LeetCode 123 ）hard

动态规划
维护四个状态

class Solution {
    
    public int maxProfit(int[] prices) {
    
        int n = prices.length;
        int buy1 = -prices[0], sell1 = 0;
        int buy2 = -prices[0], sell2 = 0;
        for (int i = 1; i < n; ++i) {
    
            buy1 = Math.max(buy1, -prices[i]);
            sell1 = Math.max(sell1, buy1 + prices[i]);
            buy2 = Math.max(buy2, sell1 - prices[i]);
            sell2 = Math.max(sell2, buy2 + prices[i]);
        }
        return sell2;
    }
}

4.买卖股票的最佳时机IV（ LeetCode 188 ）hard

动态规划
参考第三题

 class Solution {
    
        public int maxProfit(int k, int[] prices) {
    
            int n = prices.length;
            if (n < 1 || k < 1)return 0;//临界条件prices为空，或者k = 0一次股票都不能买时
            int[][] buyOrSell = new int[k][2];
            //buyOrSell[i][j] j = 0 时表示第i天结束手中持有股票的最大利润
            //buyOrSell[i][j] j = 1 时表示第i天结束手中没有持有股票的最大利润
            for (int i = 0; i < k; i++) {
    
                buyOrSell[i][0] = -prices[0];
            }
            for (int i = 1; i < n; i++) {
    
                for (int j = 0; j < k; j++) {
    
                    //当前为第j次购买股票
                    //第i天结束时手中若有股票
                    //第i结束时有股票的最大利润 = Math.max(前一天手中有第j次买入股票的最大利润，
                    // 前一天售完第j - 1次股票的最大利润 + prices[i])
                    int x = j == 0 ? 0 : buyOrSell[j - 1][1];
                    buyOrSell[j][0] = Math.max(buyOrSell[j][0],x - prices[i]);
                    buyOrSell[j][1] = Math.max(buyOrSell[j][1],buyOrSell[j][0] + prices[i]);
                }
            }

            return buyOrSell[k - 1][1];
        }
    }

5.最佳买卖股票时机含冷冻期(LeetCode 309)

动态规划
参考上面的几道题

 class Solution {
    
        public int maxProfit(int[] prices) {
    
            if (prices.length < 1)return 0;
            int buy = -prices[0];//当天结束手中有入股票
            int sellno = 0;//当天结束手中没有股票，且当天没有出售股票
            int sell = 0;//当天结束手中没有股票，且当天有出售股票
            for (int i = 1; i < prices.length; i++) {
    
                buy = Math.max(buy,sellno - prices[i]);
                sellno = Math.max(sellno,sell);
                sell = buy + prices[i];
            }
            return Math.max(sellno,sell);
        }
    }