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                合規國際互聯網加速 OSASE為企業客戶提供高速穩定SD-WAN國際加速解決方案。 廣告
                從尾到頭打印鏈表 **規則** 輸入:鏈表head 輸出: 字符串,每個元素的值 case: ``` head為null 只有一個節點 有多個節點 ``` **思路** 從頭到尾打印比較簡單,但從尾到頭打印就需要進行一些思考了 ,首先要思考能否修改鏈表,比如講鏈表進行逆序再打印,復雜度為O(n)和O(1)。本題目的要求不能修改鏈表。 思路1:申請空間,構造另外一個鏈表,使用頭插法,復雜度為O(n)和O(n) 思路2:遍歷鏈表,保存到棧中,最后出棧并打印,復雜度為O(n)和O(n) 思路3:有了棧的思路,可以使用遞歸,復雜度為O(n)和O(n) **代碼** ``` // 使用棧 public String reverse(ListNode head) { if(head == null) return ""; Stack<Integer> stack = new Stack<Integer>(); while(head != null) { stack.push(head.val); head = head.next; } StringBuilder sb = new StringBuilder(); while(!stack.isEmpty()) { int tmp = stack.pop(); sb.append(tmp); } return sb.toString(); } // 遞歸 public String reverse(ListNode head) { if(head == null) return ""; return reverse(head.next) + head.val; } ```
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