將單向連結串列按某值劃分成左邊小、中間相等、右邊大的形式（O（1）空間複雜度，穩定劃分）
阿新 • • 發佈：2018-12-24
class Solution {
private static ListNode function(ListNode head, int val) {
ListNode left = new ListNode(0);
ListNode mid = new ListNode(0);
ListNode right = new ListNode(0);
ListNode tail1 = null, tail2 = null, tail3 = null;
ListNode next;
ListNode cur = head;
while (cur != null) {
next = cur.next;
if (cur.val < val) {
append(left, tail1, cur);
tail1 = cur;
} else if (cur.val == val) {
append(mid, tail2, cur);
tail2 = cur;
} else {
append(right, tail3, cur);
tail3 = cur;
}
cur = next;
}
return merge(left.next, tail1, mid.next, tail2, right.next, tail3);
}
private static void print(ListNode head) {
while (head != null) {
System.out.println(head.val);
head = head.next;
}
}
public static void append(ListNode list, ListNode tail, ListNode node) {
if (list == null) {
return;
}
if (tail != null) {
tail.next = node;
} else {
list.next = node;
}
node.next = null;
}
public static ListNode merge(ListNode l1, ListNode tail1,
ListNode l2, ListNode tail2,
ListNode l3, ListNode tail3) {
ListNode head;
// find head
if (l1 != null) {
head = l1;
} else if (l2 != null) {
head = l2;
} else {
head = l3;
}
if (tail1 != null) {
tail1.next = l2;
}
if (tail2 != null) {
tail2.next = l3;
} else if (tail1 != null) {
tail1.next = l3;
}
tail3.next = null;
return head;
}
public static void main(String[] args) {
ListNode n1 = new ListNode(9);
ListNode n2 = new ListNode(0);
ListNode n3 = new ListNode(4);
ListNode n4 = new ListNode(5);
ListNode n5 = new ListNode(1);
n1.next = n2;
n2.next = n3;
n3.next = n4;
n4.next = n5;
ListNode head = function(n1, 3);
print(head);
}
}
class ListNode {
int val;
ListNode next;
ListNode(int x) { val = x; }
}