20190313

code/lc121.java

@@ -4,17 +4,14 @@ package code;
  * 题意：股票买卖1次，最大利润
  * 难度：Easy
  * 分类：Arryas, Dynamic Programming
- * Tips：lc122
+ * Tips：lc121, lc309, lc188, lc123, lc714
  */
 public class lc121 {
     public int maxProfit(int[] prices) {
-        int min = prices[0];if(prices.length==0)
-            return 0;
-        int res =0;
-        for (int i = 1; i < prices.length ; i++) {
-            res = Math.max(prices[i]-min,res);
-            if(prices[i]<min)
-                min = prices[i];
+        int min = Integer.MAX_VALUE, res=0;
+        for(int i=0; i<prices.length; i++){
+            min = Math.min(min, prices[i]);
+            res = Math.max(res, prices[i]-min);
         }
         return res;
     }

code/lc122.java

@@ -5,7 +5,7 @@ package code;
  * 难度：Easy
  * 分类：Array, Greedy
  * 思路：计算 prices[i] 与 prices[i-1] 的差值，把正数全加起来就行了
- * Tips：lc121, lc309
+ * Tips：lc121, lc309, lc188, lc123, lc714
  */
 public class lc122 {
     public int maxProfit(int[] prices) {

code/lc123.java

+package code;
+/*
+ * 122. Best Time to Buy and Sell Stock III
+ * 题意：买卖股票最大利润，只能买卖2次
+ * 难度：Hard
+ * 分类：Array, Dynamic Programming
+ * 思路：两种思路，第一种是分成两块，每块按lc121的算法进行计算，最后合并结果
+ *      第二种思路设置4个变量，分别为第一次买，第一次卖，第二次买，第二次卖的价格
+ * Tips：只想到了O(N^2)的方法
+ *      lc121, lc309, lc188, lc123, lc714
+ */
+public class lc123 {
+    public int maxProfit(int[] prices) {
+        int buy1 = Integer.MAX_VALUE, buy2 = Integer.MAX_VALUE, sell1 = 0, sell2 = 0;
+        for (int i = 0; i < prices.length ; i++) {
+            buy1 = Math.min(buy1, prices[i]);   //第一次购买的最低价格
+            sell1 = Math.max(sell1, prices[i] - buy1);
+            buy2 = Math.min(buy2, prices[i]-sell1); //记住第二项 prices[i]-sell1
+            sell2 = Math.max(sell2, prices[i]-buy2);    //当只购买一次时，会传递的
+        }
+        return sell2;
+    }
+
+    public int maxProfit2(int[] prices) {   //常规方法，分为两块，O(N^2)
+        int res = 0;
+        for (int i = 0; i <prices.length ; i++) {
+            int res1 = Math.max(0, helper(prices,0,i));
+            int res2 = Math.max(0, helper(prices, i, prices.length));
+            res = Math.max(res, res1+res2);
+        }
+        return res;
+    }
+    public int helper(int[] prices, int begin, int end) {
+        int min = Integer.MAX_VALUE, res=0;
+        for(int i=begin; i<end; i++){
+            min = Math.min(min, prices[i]);
+            res = Math.max(res, prices[i]-min);
+        }
+        return res;
+    }
+}

code/lc188.java

+package code;
+/*
+ * 188. Best Time to Buy and Sell Stock IV
+ * 题意：买卖股票最大利润，可以买卖k次
+ * 难度：Hard
+ * 分类：Dynamic Programming
+ * 思路：二维dp, dp[i][j] 表示i次交易，在数组prices[0~j]上的最大利润
+ *      dp[i][j] = Max( dp[i][j-1], dp[i-1][jj]+prices[j]-prices[jj] )   { jj in range of [0, j-1] }
+ *               = Max( dp[i][j-1], prices[j]+ max(dp[i-1][jj]-prices[jj]) ) 转化为这一步，少了一层循环
+ *      dp[0][j] = 0;  dp[i][0] = 0;
+ * Tips：lc121, lc309, lc188, lc123, lc714
+ */
+public class lc188 {
+    public int maxProfit(int k, int[] prices) {
+        if(prices.length==0) return 0;
+        int n = prices.length;
+        //if k >= n/2, then you can make maximum number of transactions.
+        if (k >=  n/2) {
+            int maxPro = 0;
+            for (int i = 1; i < n; i++) {
+                if (prices[i] > prices[i-1])
+                    maxPro += prices[i] - prices[i-1];
+            }
+            return maxPro;
+        }
+
+        int[][] dp = new int[k+1][prices.length];
+        for (int i = 1; i <= k ; i++) {
+            int localMax = -prices[0];
+            for (int j = 1; j < prices.length ; j++) {  //jj的计算和这一维合并，总的复杂度是二次方而不是三次
+                dp[i][j] = Math.max(dp[i][j-1], prices[j]+localMax);
+                localMax = Math.max(localMax, dp[i-1][j]-prices[j]);
+            }
+        }
+        return dp[k][prices.length-1];
+    }
+}

code/lc309.java

@@ -9,7 +9,7 @@ package code;
  *      sell[i] = max( sell[i-1], buy[i-1]+price[i] )
  *      空间压缩以后时间是O(n)，空间是O(1)
  * Tips：https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-cooldown/discuss/75931/Easiest-JAVA-solution-with-explanations
- *  lc122
+ *      lc121, lc309, lc188, lc123, lc714
  */
 public class lc309 {
     public int maxProfit(int[] prices) {

code/lc714.java

+package code;
+/*
+ * 714. Best Time to Buy and Sell Stock with Transaction Fee
+ * 题意：买卖股票，不限次数，但有交易费用，求最大利润
+ * 难度：Medium
+ * 分类：Array, Dynamic Programming, Greedy
+ * 思路：和309思路一致，每次卖出的时候减去交易费用
+ *      buy[i] = max( buy[i-1], sell[i-1]-price[i] )
+ *      sell[i] = max( sell[i-1], buy[i-1]+price[i]-2 )
+ * Tips：
+ *      总结一下买卖股票的问题  交易1次，2次，任意多次都可在O(N)完成交易
+ *      交易k次时，时间复杂度为O(NM),M为交易次数
+ *      冷却时间和交易费用的解法一致，都是分买和卖两个状态，进行dp
+ */
+public class lc714 {
+    public int maxProfit(int[] prices, int fee) {
+        if(prices.length==0) return 0;
+        int b1 = -prices[0];
+        int s1=0, b = 0, s = 0;
+        for (int i = 0; i < prices.length ; i++) {
+            b = Math.max(b1, s1-prices[i]);
+            s = Math.max(s1, b1+prices[i]-fee);
+            s1 = s;
+            b1 = b;
+        }
+        return Math.max(s,b);
+    }
+}

readme.md

@@ -45,7 +45,9 @@ LeetCode 指南
 | 021 [Java](./code/lc21.java)
 | 022 [Java](./code/lc22.java)
 | 023 [Java](./code/lc23.java)
+| 024 [Java](./code/lc24.java)
 | 026 [Java](./code/lc26.java)
+| 027 [Java](./code/lc27.java)
 | 028 [Java](./code/lc28.java)
 | 029 [Java](./code/lc29.java)
 | 031 [Java](./code/lc31.java)
@@ -96,6 +98,7 @@ LeetCode 指南
 | 114 [Java](./code/lc114.java)
 | 116 [Java](./code/lc116.java)
 | 118 [Java](./code/lc118.java)
+| 120 [Java](./code/lc120.java)
 | 121 [Java](./code/lc121.java)
 | 122 [Java](./code/lc122.java)
 | 124 [Java](./code/lc124.java)
@@ -138,6 +141,7 @@ LeetCode 指南
 | 208 [Java](./code/lc208.java)
 | 210 [Java](./code/lc210.java)
 | 212 [Java](./code/lc212.java)
+| 213 [Java](./code/lc213.java)
 | 215 [Java](./code/lc215.java)
 | 217 [Java](./code/lc217.java)
 | 218 [Java](./code/lc215.java)