題目
輸入一個鏈表的頭節點,從尾到頭反過來返回每個節點的值(用數組返回)。
示例 1:
輸入:head = [1,3,2]
輸出:[2,3,1]
限制:
0 <= 鏈表長度 <= 10000
第一次
//使用棧的特性先進後出
//複雜度O(n)
public int[] reversePrint(ListNode head) {
Stack<ListNode> stack = new Stack<ListNode>();
ListNode temp = head;
while(temp!= null){
stack.push(temp);
temp = temp.next;
}
int[] result = new int[stack.size()];
for(int i = 0 ; i< stack.size();i ++ ){
result[i] = stack.pop().val;
}
return result;
}
//自己腦瓜子有點問題
public int[] reversePrint(ListNode head) {
ListNode[] result = new ListNode[];
ListNode temp = head;
int count = 0;
while(temp != null ){
result[count] = temp;
temp = temp.next;
count++;
}
int[] resultV2 = new int[count];
for(int i = 0; i<count; i++){
resultV2[count] = result[i].val;
count--;
}
return resultV2;
}
第三種解決方案
//純O(n)的複雜度
class Solution {
public int[] reversePrint(ListNode head) {
ListNode temp = head;
int count = 0;
while(temp != null ){
temp = temp.next;
count++;
}
int[] resultV2 = new int[count];
temp = head;
for(int i = count-1; i>=0; i--){
resultV2[i] = temp.val;
temp = temp.next;
}
return resultV2;
}
}