标签：-- interview temp pre append int waiting 直接 tmp
第二轮最后interviewer 直接说you are done for today
1st round:-google 1point3acres
you have 2 available operations, /3 and *2, given an integer n, what is the minimum steps to get n from 1 using those 2 ops.
2nd round:
stock prices
The interviewer in 2nd round said and I quote "many interviewers dropped out so they found me sorry keep you waiting"
pre 和end 的位置
class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
ListNode h = head;
int len = 0;
while (h != null) {
len++;
h = h.next;
}
if (len < 2 || k == 1 || len < k) {
return head;
}
int m = len / k;
ListNode end = null;
ListNode newHead = null;
ListNode pre = null;
while (m > 0) {
int cnt = k;
ListNode tail = null;
end = head;
while (cnt > 0) {
ListNode temp = head.next;
head.next = tail;
tail = head;
head = temp;
cnt--;
}
if (m == len / k) {
newHead = tail;
} else {
pre.next = tail;
}
pre = end;
m--;
}
end.next = head;
return newHead;
}
}
recursion
public ListNode reverseKGroup(ListNode head, int k) {
ListNode curr = head;
int count = 0;
while (curr != null && count != k) { // find the k+1 node
curr = curr.next;
count++;
}
if (count == k) { // if k+1 node is found
curr = reverseKGroup(curr, k); // reverse list with k+1 node as head
// head - head-pointer to direct part,
// curr - head-pointer to reversed part;
while (count-- > 0) { // reverse current k-group:
ListNode tmp = head.next; // tmp - next head in direct part
head.next = curr; // preappending "direct" head to the reversed list
curr = head; // move head of reversed part to a new node
head = tmp; // move "direct" head to the next node in direct part
}
head = curr;
}
return head;
}
标签：-- interview temp pre append int waiting 直接 tmp
原文地址：http://www.cnblogs.com/apanda009/p/7858970.html
