From c4bc3197502473c5f14be392157bb83ddaa2dbeb Mon Sep 17 00:00:00 2001
From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com>
Date: Fri, 5 May 2023 21:39:52 -0500
Subject: [PATCH] =?UTF-8?q?Update=200018.=E5=9B=9B=E6=95=B0=E4=B9=8B?=
 =?UTF-8?q?=E5=92=8C.md?=
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---
 problems/0018.四数之和.md | 33 +++++++++++----------------------
 1 file changed, 11 insertions(+), 22 deletions(-)

diff --git a/problems/0018.四数之和.md b/problems/0018.四数之和.md
index 4fee8358..a4d41d9b 100644
--- a/problems/0018.四数之和.md
+++ b/problems/0018.四数之和.md
@@ -252,36 +252,25 @@ class Solution(object):
         :type target: int
         :rtype: List[List[int]]
         """
-        # use a dict to store value:showtimes
-        hashmap = dict()
-        for n in nums:
-            if n in hashmap:
-                hashmap[n] += 1
-            else: 
-                hashmap[n] = 1
+        # 创建一个字典来存储输入列表中每个数字的频率
+        freq = {}
+        for num in nums:
+            freq[num] = freq.get(num, 0) + 1
         
-        # good thing about using python is you can use set to drop duplicates.
+        # 创建一个集合来存储最终答案，并遍历4个数字的所有唯一组合
         ans = set()
-        # ans = []  # save results by list()
         for i in range(len(nums)):
             for j in range(i + 1, len(nums)):
                 for k in range(j + 1, len(nums)):
                     val = target - (nums[i] + nums[j] + nums[k])
-                    if val in hashmap:
-                        # make sure no duplicates.
+                    if val in freq:
+                        # 确保没有重复
                         count = (nums[i] == val) + (nums[j] == val) + (nums[k] == val)
-                        if hashmap[val] > count:
-                          ans_tmp = tuple(sorted([nums[i], nums[j], nums[k], val]))
-                          ans.add(ans_tmp)
-                          # Avoiding duplication in list manner but it cause time complexity increases
-                          # if ans_tmp not in ans:  
-                          #     ans.append(ans_tmp)
-                        else:
-                            continue
-        return list(ans) 
-        # if used list() to save results, just 
-        # return ans
+                        if freq[val] > count:
+                            ans.add(tuple(sorted([nums[i], nums[j], nums[k], val])))
         
+        return [list(x) for x in ans]
+
 ```
 
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