From ac49e879e0c913a65e6a3955b066aafef1641bcd Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E6=9E=81=E5=AE=A2=E5=AD=A6=E4=BC=9F?= <qxuwei@yeah.net>
Date: Thu, 2 Sep 2021 13:29:07 +0800
Subject: [PATCH] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=20=E7=AC=AC454=E9=A2=98.?=
 =?UTF-8?q?=E5=9B=9B=E6=95=B0=E7=9B=B8=E5=8A=A0II=20Swift=E7=89=88?=
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---
 problems/0454.四数相加II.md | 29 +++++++++++++++++++++++++++--
 1 file changed, 27 insertions(+), 2 deletions(-)

diff --git a/problems/0454.四数相加II.md b/problems/0454.四数相加II.md
index 4e61dc2f..1c3f45e7 100644
--- a/problems/0454.四数相加II.md
+++ b/problems/0454.四数相加II.md
@@ -220,8 +220,33 @@ var fourSumCount = function(nums1, nums2, nums3, nums4) {
 };
 ```
 
-
-
+Swift：
+```swift
+func fourSumCount(_ nums1: [Int], _ nums2: [Int], _ nums3: [Int], _ nums4: [Int]) -> Int {
+    // key:a+b的数值，value:a+b数值出现的次数
+    var map = [Int: Int]()
+    // 遍历nums1和nums2数组，统计两个数组元素之和，和出现的次数，放到map中
+    for i in 0 ..< nums1.count {
+        for j in 0 ..< nums2.count {
+            let sum1 = nums1[i] + nums2[j]
+            map[sum1] = (map[sum1] ?? 0) + 1
+        }
+    }
+    // 统计a+b+c+d = 0 出现的次数
+    var res = 0
+    // 在遍历大num3和num4数组，找到如果 0-(c+d) 在map中出现过的话，就把map中key对应的value也就是出现次数统计出来。
+    for i in 0 ..< nums3.count {
+        for j in 0 ..< nums4.count {
+            let sum2 = nums3[i] + nums4[j]
+            let other = 0 - sum2
+            if map.keys.contains(other) {
+                res += map[other]!
+            }
+        }
+    }
+    return res
+}
+```
 
 -----------------------
 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)