Merge pull request #2800 from zrzhit/master

增加Swift版本  1. 面试题02.07.链表相交  2.右旋字符串

problems/kamacoder/0055.右旋字符串.md

@@ -350,7 +350,29 @@ function reverseStr(s, start, end) {
 
 
 ### Swift:
+```swift
+func rotateWords(_ s: String, _ k: Int) -> String {
+    var chars = Array(s)
+    // 先反转整体
+    reverseWords(&chars, start: 0, end: s.count - 1)
+    // 反转前半段
+    reverseWords(&chars, start: 0, end: k - 1)
+    // 反转后半段
+    reverseWords(&chars, start: k, end: s.count - 1)
+    return String(chars)
+}
 
+// 反转start...end 的字符数组
+func reverseWords(_ chars: inout [Character], start: Int, end: Int) {
+    var left = start
+    var right = end
+    while left < right, right < chars.count {
+        (chars[left], chars[right]) = (chars[right], chars[left])
+        left += 1
+        right -= 1
+    }
+}
+```
 
 
 ### PHP：

problems/面试题02.07.链表相交.md

@@ -535,6 +535,45 @@ public ListNode GetIntersectionNode(ListNode headA, ListNode headB)
 }
 ```
 
+### Swift：
+```swift
+func getIntersectionNode(_ headA: ListNode?, _ headB: ListNode?) -> ListNode? {
+    var lenA = 0
+    var lenB = 0
+    var nodeA = headA
+    var nodeB = headB
+    // 计算链表A和链表B的长度
+    while nodeA != nil {
+        lenA += 1
+        nodeA = nodeA?.next
+    }
+    while nodeB != nil {
+        lenB += 1
+        nodeB = nodeB?.next
+    }
+    // 重置指针
+    nodeA = headA
+    nodeB = headB
+    // 如果链表A更长，让它先走lenA-lenB步
+    if lenA > lenB {
+        for _ in 0..<(lenA - lenB) {
+            nodeA = nodeA?.next
+        }
+    } else if lenB > lenA {
+        // 如果链表B更长，让它先走lenB-lenA步
+        for _ in 0..<(lenB - lenA) {
+            nodeB = nodeB?.next
+        }
+    }
+    // 同时遍历两个链表，寻找交点
+    while nodeA !== nodeB {
+        nodeA = nodeA?.next
+        nodeB = nodeB?.next
+    }
+    return nodeA
+}
+```
+
 
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