Problem
Reverse a singly linked list.Analysis
- 这道题就是要把相邻两个node的指针指向反一反,反的过程当中要keep track of next node,不然这时候 node.next 变成前一个节点了
- 所以我们采用两个指针,prev 和 curr 来记录相邻的两个节点
- 反之前,先记录 curr.next, 即 tmp = curr.next
- 然后可以反了,curr.next = prev
- 然后把prev 和 curr往右移,继续去处理还没反的nodes
- prev = curr
- curr = tmp
- Time: O(n)
- Space: O(1)
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