思路:用兩對前驅和後繼節點,分别比較目前節點的前驅和後繼以及最小值界定啊的前驅和後繼。
周遊完整個連結清單,删除最小值節點即可。
struct node {
int val;
node *next;
};
void deleteMin(node *head) {
node *p = head, *q = head->next;
node *premin = head, *pmin = head->next;
while (q) {
if (q->val < pmin->val) {
pmin = q;
premin = p;
p = p->next, q = q->next;
} else {
p = p->next;
q = q->next;
}
node *del = pmin;
premin->next = pmin->next;
delete del;
}
}
複制