Problem: 206. 反转链表
解析
这个题干想很费脑子，多个指针变来变去可记不住，浪费一个小时也没调对，还是得画下图，一步一步规划代码，类似数形结合思想，这叫码形结合思想
码形结合思想
开头额外处理，剩下的节点循环处理即可
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class Solution {
public:
ListNode *reverseList(ListNode *head) {
if (head == nullptr) {
return nullptr;
}
ListNode *node = head;
ListNode *pre;
ListNode *nex;
pre = node;
if (node->next == nullptr) {
return node;
}
node = node->next;
nex = node->next;
node->next = pre;
pre->next = nullptr;
pre = node;
node = nex;
while (node != nullptr) {
nex = node->next;
node->next = pre;
pre = node;
node = nex;
}
return pre;
}
};
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简化代码
上个答案对不必要逻辑和变量的缩减
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class Solution {
public:
ListNode *reverseList(ListNode *head) {
ListNode *node = head;
ListNode *pre = nullptr;
ListNode *nex;
pre = node;
if (node->next == nullptr) {
return node;
}
node = node->next;
nex = node->next;
node->next = pre;
pre->next = nullptr;
pre = node;
node = nex;
while (node != nullptr) {
nex = node->next;
node->next = pre;
pre = node;
node = nex;
}
return pre;
}
};
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第一次做的错误答案
错误答案，第一遍给看成循环移动链表的了，好好审题，浪费时间了
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| class Solution {
public:
ListNode *reverseList(ListNode *head) {
if (head == nullptr) {
return nullptr;
}
ListNode *node1 = head;
ListNode *tail = head;
while (tail->next != nullptr) {
tail = tail->next;
}
ListNode *node2 = tail;
while (node1->next != tail) {
node2->next = node1;
node2 = node2->next;
node1 = node1->next;
}
node2->next = node1;
node1->next = nullptr;
return tail;
}
};
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