Description:
Given the head of a linked list, we repeatedly delete consecutive sequences of nodes that sum to 0 until there are no such sequences.
After doing so, return the head of the final linked list. You may return any such answer.
Input: head = [1,2,-3,3,1]
Output: [3,1]
Note: The answer [1,2,1] would also be accepted.
The following answer's complexity is O(n) in both time and space, and is very easy to understand:leetcode-solution-1171
if P1 + P2 = P1 + P2 + P3 + P4 + P5, it means the sum of P3 + P4 + P5 is 0. As the sum is consecutive sequences of nodes. The specular attention need to be paid is that the sum of linked list is 0, and a 0 node is added before the head which can track linked list and trackle this situation.
fun removeZeroSumSublists(head: ListNode?): ListNode? {
// 1. the head need to remove; 2.the sum of linked list is 0;
val preHead = ListNode(0)
preHead.next = head
// including the sum of whole linked list is 0.
var cur: ListNode? = preHead
val map = mutableMapOf<Int, ListNode>()
var sum = 0
while (cur != null) {
sum += cur.`val`
map[sum] = cur
cur = cur.next
}
sum = 0
cur = preHead
while (cur != null) {
sum += cur.`val`
cur.next = map[sum]?.next
cur = cur.next
}
return preHead.next
}
