Merge pull request #705 from qxuewei/master

添加 第454题.四数相加II Swift版本

problems/0383.赎金信.md

@@ -266,6 +266,28 @@ var canConstruct = function(ransomNote, magazine) {
 };
 ```
 
+Swift：
+```swift
+func canConstruct(_ ransomNote: String, _ magazine: String) -> Bool {
+    var record = Array(repeating: 0, count: 26);
+    let aUnicodeScalarValue = "a".unicodeScalars.first!.value
+    for unicodeScalar in magazine.unicodeScalars {
+        // 通过record 记录 magazine 里各个字符出现的次数
+        let idx: Int = Int(unicodeScalar.value - aUnicodeScalarValue)
+        record[idx] += 1
+    }
+    for unicodeScalar in ransomNote.unicodeScalars {
+        // 遍历 ransomNote,在record里对应的字符个数做 -- 操作
+        let idx: Int = Int(unicodeScalar.value - aUnicodeScalarValue)
+        record[idx] -= 1
+        // 如果小于零说明在magazine没有
+        if record[idx] < 0 {
+            return false
+        }
+    }
+    return true
+}
+```
 
 
 -----------------------

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
+}
+```
 
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 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)