6.從尾到頭打印鏈表
題目描述
輸入一個鏈表,從尾到頭打印鏈表每個節點的值。
題目很經典,因此本文用三種方法來處理。(嚴格來說算2種)
思路1
先順序輸出鏈表數據,然後翻轉輸出的結果。
/**
* struct ListNode {
* int val;
* struct ListNode *next;
* ListNode(int x) :
* val(x), next(NULL) {
* }
* };
*/
class Solution {
public:
vector<int> printListFromTailToHead(ListNode* head) {
if (head == NULL) return {};
vector<int> ret;
ListNode* pRead = head;
while(pRead != NULL) {
ret.push_back(pRead->val);
pRead = pRead->next;
}
std::reverse(ret.begin(), ret.end());
return ret;
}
};
從頭到位遍歷一遍數據,因此時間複雜度:O(n)
空間複雜度:O(n)
思路2 棧 非遞歸
/**
* struct ListNode {
* int val;
* struct ListNode *next;
* ListNode(int x) :
* val(x), next(NULL) {
* }
* };
*/
class Solution {
public:
vector<int> printListFromTailToHead(ListNode* head) {
if (head == NULL) return {};
std::stack<ListNode*> sk;
ListNode* pRead = head;
while (pRead != NULL) {
sk.push(pRead);
pRead = pRead->next;
}
vector<int> ret;
while (!sk.empty()) {
ret.push_back(sk.top()->val);
sk.pop();
}
return ret;
}
};
思路3 棧 遞歸
class Solution {
public:
vector<int> printListFromTailToHead(ListNode* head) {
if (head == NULL) return {};
vector<int> ret;
ListNode* pRead = head;
help(ret, pRead);
return ret;
}
void help(vector<int>& ret, ListNode* node) {
if (node == NULL) return;
help(ret, node->next);
ret.push_back(node->val);
}
};