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題目描述
輸入兩個單調遞增的鏈表,輸出兩個鏈表合成後的鏈表,當然我們需要合成後的鏈表滿足單調不減規則。
遞歸版:
/*
public class ListNode {
int val;
ListNode next = null;
ListNode(int val) {
this.val = val;
}
}*/
public class Solution {
public ListNode Merge(ListNode list1,ListNode list2) {
if (list1 == null) return list2;
if (list2 == null) return list1;
if (list1.val <= list2.val) {
list1.next = Merge(list1.next, list2);
return list1;
} else {
list2.next = Merge(list1, list2.next);
return list2;
}
}
}
非遞歸版:(外排)
需要注意如何處理頭結點爲空的情況
/*
public class ListNode {
int val;
ListNode next = null;
ListNode(int val) {
this.val = val;
}
}*/
public class Solution {
public ListNode Merge(ListNode list1,ListNode list2) {
if (list1 == null) return list2;
if (list2 == null) return list1;
ListNode res = null;
ListNode cur = res;
while (list1 != null && list2 != null) {
if (list1.val <= list2.val) {
if (res == null) {
res = list1;
cur = list1;
} else {
cur.next = list1;
cur = cur.next; // 把下一個節點作爲下一次的當前節點
}
list1 = list1.next;
} else {
if (res == null) {
res = list2;
cur = list2;
} else {
cur.next = list2;
cur = cur.next; // 把下一個節點作爲下一次的當前節點
}
list2 = list2.next;
}
}
// list1沒越界, list2必越界
if (list1 != null) {
cur.next = list1;
}
// list2沒越界,list1必越界
if (list2 != null) {
cur.next = list2;
}
return res;
}
}