Merge pull request doocs#298 from gengyuanzhe/master

0740 Delete and Earn新增Java解

Author: yanglbme

solution/0700-0799/0740.Delete and Earn/README.md

@@ -43,8 +43,24 @@
 ## 解法
 <!-- 这里可写通用的实现逻辑 -->
 
-
 <!-- tabs:start -->
+核心思路: **一个数字要么不选，要么全选**
+
+首先计算出每个数字的总和 sums，并维护两个 dp 数组：select 和 nonSelect
+
+- sums[i] 代表值为 i 的元素总和
+- select[i] 代表如果选数字 i，从 0 处理到 i 的最大和
+- nonSelect[i] 代表如果不选数字 i，从 0 处理到 i 的最大和
+
+那么我们有以下逻辑：
+
+- 如果选 i，那么 i-1 肯定不能选；
+- 如果不选 i，那么 i-1 选不选都可以，因此我们选择其中较大的选法
+
+``` java
+select[i] = nonSelect[i-1] + sums[i];
+nonSelect[i] = Math.max(select[i-1], nonSelect[i-1]);
+```
 
 ### **Python3**
 <!-- 这里可写当前语言的特殊实现逻辑 -->
@@ -57,7 +73,29 @@
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class Solution {
+    public int deleteAndEarn(int[] nums) {
+        if (nums.length == 0) {
+            return 0;
+        }
+
+        int[] sums = new int[10010];
+        int[] select = new int[10010];
+        int[] nonSelect = new int[10010];
+
+        int maxV = 0;
+        for (int x : nums) {
+            sums[x] += x;
+            maxV = Math.max(maxV, x);
+        }
+
+        for (int i = 1; i <= maxV; i++) {
+            select[i] = nonSelect[i - 1] + sums[i];
+            nonSelect[i] = Math.max(select[i - 1], nonSelect[i - 1]);
+        }
+        return Math.max(select[maxV], nonSelect[maxV]);
+    }
+}
 ```
 
 ### **...**

solution/0700-0799/0740.Delete and Earn/README_EN.md

@@ -86,8 +86,24 @@ Then, delete 3 again to earn 3 points, and 3 again to earn 3 points.
 
 ## Solutions
 
-
 <!-- tabs:start -->
+Intuition: **If we take a number, we will take all of the copies of it**.
+
+First calculate the sum of each number as **sums**, and keep updating two dp arrays: **select** and **nonSelect**
+
+- sums[i] represents the sum of elements whose value is i;
+- select[i] represents the maximum sum of processing from 0 to i if the number i is selected;
+- nonSelect[i] represents the maximum sum of processing from 0 to i if the number i is not selected;
+
+Then we have the following conclusions:
+
+- If i is selected, then i-1 must not be selected;
+- If you do not choose i, then i-1 can choose or not, so we choose the larger one;
+
+```java
+select[i] = nonSelect[i-1] + sums[i];
+nonSelect[i] = Math.max(select[i-1], nonSelect[i-1]);
+```
 
 ### **Python3**
 
@@ -98,7 +114,29 @@ Then, delete 3 again to earn 3 points, and 3 again to earn 3 points.
 ### **Java**
 
 ```java
-
+class Solution {
+    public int deleteAndEarn(int[] nums) {
+        if (nums.length == 0) {
+            return 0;
+        }
+
+        int[] sums = new int[10010];
+        int[] select = new int[10010];
+        int[] nonSelect = new int[10010];
+
+        int maxV = 0;
+        for (int x : nums) {
+            sums[x] += x;
+            maxV = Math.max(maxV, x);
+        }
+
+        for (int i = 1; i <= maxV; i++) {
+            select[i] = nonSelect[i - 1] + sums[i];
+            nonSelect[i] = Math.max(select[i - 1], nonSelect[i - 1]);
+        }
+        return Math.max(select[maxV], nonSelect[maxV]);
+    }
+}
 ```
 
 ### **...**

solution/0700-0799/0740.Delete and Earn/Solution.java

@@ -0,0 +1,23 @@
+public class Solution {
+    public int deleteAndEarn(int[] nums) {
+        if (nums.length == 0) {
+            return 0;
+        }
+
+        int[] sums = new int[10010];
+        int[] select = new int[10010];
+        int[] nonSelect = new int[10010];
+
+        int maxV = 0;
+        for (int x : nums) {
+            sums[x] += x;
+            maxV = Math.max(maxV, x);
+        }
+
+        for (int i = 1; i <= maxV; i++) {
+            select[i] = nonSelect[i - 1] + sums[i];
+            nonSelect[i] = Math.max(select[i - 1], nonSelect[i - 1]);
+        }
+        return Math.max(select[maxV], nonSelect[maxV]);
+    }
+}