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Commit e816450

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feat: update solutions to lc problem: No.0740. Delete and Earn
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-25
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‎README.md‎

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- [使用最小花费爬楼梯](./solution/0700-0799/0746.Min%20Cost%20Climbing%20Stairs/README.md)
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- [打家劫舍](./solution/0100-0199/0198.House%20Robber/README.md)
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- [打家劫舍 II](./solution/0200-0299/0213.House%20Robber%20II/README.md)
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- [删除并获得点数](./solution/0700-0799/0740.Delete%20and%20Earn/README.md)
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- [接雨水](./solution/0000-0099/0042.Trapping%20Rain%20Water/README.md)
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- [最大子序和](./solution/0000-0099/0053.Maximum%20Subarray/README.md)
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- [礼物的最大价值](./lcof/面试题47.%20礼物的最大价值/README.md)

‎README_EN.md‎

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@@ -158,6 +158,7 @@ Complete solutions to [LeetCode](https://leetcode.com/problemset/all/), [LCOF](h
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- [Min Cost Climbing Stairs](./solution/0700-0799/0746.Min%20Cost%20Climbing%20Stairs/README_EN.md)
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- [House Robber](./solution/0100-0199/0198.House%20Robber/README_EN.md)
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- [House Robber II](./solution/0200-0299/0213.House%20Robber%20II/README_EN.md)
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- [Delete and Earn](./solution/0700-0799/0740.Delete%20and%20Earn/README_EN.md)
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- [Trapping Rain Water](./solution/0000-0099/0042.Trapping%20Rain%20Water/README_EN.md)
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- [Maximum Subarray](./solution/0000-0099/0053.Maximum%20Subarray/README_EN.md)
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- [Minimum Path Sum](./solution/0000-0099/0064.Minimum%20Path%20Sum/README_EN.md)

‎solution/0100-0199/0198.House Robber/README.md‎

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@@ -86,22 +86,14 @@ public:
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class Solution {
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public:
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int rob(vector<int>& nums) {
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int n;
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if ((n = nums.size()) == 0) return 0;
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return robRange(nums, 0, n - 1);
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}
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private:
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int robRange(vector<int>& nums, int start, int end) {
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if (end - start == 0) return nums[start];
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int pre = 0;
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int cur = nums[start];
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for (int i = start + 1; i < end + 1; ++i) {
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int t = max(pre + nums[i], cur);
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pre = cur;
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cur = t;
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int n = nums.size();
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int a = 0, b = nums[0];
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for (int i = 1; i < n; ++i) {
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int c = max(nums[i] + a, b);
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a = b;
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b = c;
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}
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return cur;
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return b;
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}
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};
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```

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

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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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核心思路: **一个数字要么不选,要么全选**
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首先计算出每个数字的总和 sums,并维护两个 dp 数组:select 和 nonSelect
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- sums[i] 代表值为 i 的元素总和
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- select[i] 代表如果选数字 i,从 0 处理到 i 的最大和
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- nonSelect[i] 代表如果不选数字 i,从 0 处理到 i 的最大和
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- `sums[i]` 代表值为 i 的元素总和
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- `select[i]` 代表如果选数字 i,从 0 处理到 i 的最大和
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- `nonSelect[i]` 代表如果不选数字 i,从 0 处理到 i 的最大和
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那么我们有以下逻辑:
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nonSelect[i] = Math.max(select[i-1], nonSelect[i-1]);
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```
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int deleteAndEarn(vector<int>& nums) {
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vector<int> vals(10010);
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for (int& num : nums) {
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vals[num] += num;
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}
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return rob(vals);
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}
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int rob(vector<int>& nums) {
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int a = 0, b = nums[0];
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for (int i = 1; i < nums.size(); ++i) {
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int c = max(nums[i] + a, b);
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a = b;
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b = c;
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}
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return b;
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}
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};
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```
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### **...**
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```

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

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## Solutions
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<!-- tabs:start -->
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Intuition: **If we take a number, we will take all of the copies of it**.
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First calculate the sum of each number as **sums**, and keep updating two dp arrays: **select** and **nonSelect**
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- sums[i] represents the sum of elements whose value is i;
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- select[i] represents the maximum sum of processing from 0 to i if the number i is selected;
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- nonSelect[i] represents the maximum sum of processing from 0 to i if the number i is not selected;
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- `sums[i]` represents the sum of elements whose value is i;
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- `select[i]` represents the maximum sum of processing from 0 to i if the number i is selected;
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- `nonSelect[i]` represents the maximum sum of processing from 0 to i if the number i is not selected;
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Then we have the following conclusions:
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nonSelect[i] = Math.max(select[i-1], nonSelect[i-1]);
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```
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<!-- tabs:start -->
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### **Python3**
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```python
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int deleteAndEarn(vector<int>& nums) {
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vector<int> vals(10010);
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for (int& num : nums) {
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vals[num] += num;
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}
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return rob(vals);
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}
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int rob(vector<int>& nums) {
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int a = 0, b = nums[0];
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for (int i = 1; i < nums.size(); ++i) {
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int c = max(nums[i] + a, b);
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a = b;
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b = c;
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}
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return b;
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}
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};
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```
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### **...**
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```
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class Solution {
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public:
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int deleteAndEarn(vector<int>& nums) {
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vector<int> vals(10010);
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for (int& num : nums) {
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vals[num] += num;
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}
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return rob(vals);
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}
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int rob(vector<int>& nums) {
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int a = 0, b = nums[0];
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for (int i = 1; i < nums.size(); ++i) {
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int c = max(nums[i] + a, b);
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a = b;
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b = c;
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}
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return b;
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}
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};

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