From 60d813e02c117b3bbe383298d64f5a6d74265e42 Mon Sep 17 00:00:00 2001
From: mahaozhong <1831866525@qq.com>
Date: Fri, 23 Apr 2021 11:00:16 +0800
Subject: [PATCH] LeetCode-368

---
 368/code.cpp    | 38 ++++++++++++++++++++++++++++++++++++++
 368/solution.md |  5 +++++
 2 files changed, 43 insertions(+)
 create mode 100644 368/code.cpp
 create mode 100644 368/solution.md

diff --git a/368/code.cpp b/368/code.cpp
new file mode 100644
index 0000000..4eea540
--- /dev/null
+++ b/368/code.cpp
@@ -0,0 +1,38 @@
+class Solution {
+public:
+    // 时间复杂度：O(n^2)
+    // 其实只需要考虑一个整除的方向即可，即使前面较小的元素集合不正确，但是最终答案是正确的
+    vector<int> largestDivisibleSubset(vector<int>& nums) {
+        int len = nums.size();
+        if (len == 1) {
+            return nums;
+        }
+        // 方便进行dp，保证一个数整除比他小的数时，下标不会超过当前元素
+        sort(nums.begin(), nums.end());
+        unordered_map<int, vector<int>> mp;
+        mp.clear();
+        for (int i = 0; i < len; i++) {
+            vector<int> temp = {nums[i]};
+            mp[i] = temp;
+            for (int j = 0; j < i; j++) {
+                if (nums[i] % nums[j] == 0) {
+                    // 可以进行转移
+                    if (mp[j].size() + 1 > mp[i].size()) {
+                        temp = mp[j];
+                        temp.emplace_back(nums[i]);
+                        mp[i] = temp;
+                    }
+                }
+            }
+        }
+        int solve = 0;
+        vector<int> ans;
+        for (auto& it : mp) {
+            if ((it).second.size() > solve) {
+                solve = (it).second.size();
+                ans = (it).second;
+            }
+        }
+        return ans;
+    }
+};
diff --git a/368/solution.md b/368/solution.md
new file mode 100644
index 0000000..7fc5903
--- /dev/null
+++ b/368/solution.md
@@ -0,0 +1,5 @@
+#### 动态规划
+* 原数组sort排序，保证了前面的数可以被后面的数整除，捋清乱序情况下数的整除关系
+* 整除具有传递性，假设三个数a，b，c满足b%a=0 和c%b=0，那么必然存在c%a=0
+* 根据上述传递性，我们可以写出转移方程，默认情况下dp[i] = {nums[i],} ，遍历j\<i，如果发现nums[i]%nums[j]=0，那么说明当前状态i可由状态j转移而来，更新状态i所对应的解即可
+* 时间复杂度：O(nlogn + n^2 + n) = O(n^2)