## 原题

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:

Given `1->2->3->4->5->NULL`, m = 2 and n = 4,

return `1->4->3->2->5->NULL`.

Note:
Given m, n satisfy the following condition:
1 ≤ mn ≤ length of list.

## 程序范例

[code lang=”cpp”]
/**
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* reverseBetween(ListNode* head, int m, int n) {
int data[10000] = {};
for (int i = 0; i < n; i++) { data[i] = chead->val;
}

int mid = (m + n)>>1;
for (int left = m – 1; left <= mid – 1; left++) {
int right = (n – 1) – (left – (m – 1));
if (right <= left) {
// while up to the mid index
break;
}
int tmp = data[left];
data[left] = data[right];
data[right] = tmp;
}

for (int i = 0; i < n; i++) {
std::cout<<data[i]<<std::endl;
}

for (int i = 0; i < n; i++) { if (i >= m – 1) {
}
}

}
};
[/code]

## 时间复杂度

O(n^3)

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