leetcode21_合併兩個有序鏈表_迭代+遞歸

一. 遞歸

1. 兩個鏈表較小的元素與剩下的元素merge併合並.

2. 時間複雜度O(n+m). 空間也爲O(n+m). 有n+m個幀棧.

class Solution {
public:
    //函數mergeTwoLists的作用是返回合併好的鏈表的頭結點.
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        //遞歸終止條件.
        if(l1==NULL) return l2;
        if(l2==NULL) return l1;
        if(l1->val<=l2->val) {
            l1->next = mergeTwoLists(l1->next, l2);
            return l1;
        }
        else {
            l2->next = mergeTwoLists(l1, l2->next);
            return l2;
        }
    }
};

二. 迭代方法.

1. 直接指向了l1和l2,類似穿針引線,這樣空間複雜度就爲O(1).

class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        //構造啞結點作爲返回鏈表的頭結點.
        ListNode* dummy = new ListNode(0);
        ListNode* cur = dummy;
        while(l1!=NULL && l2!=NULL) {
            if(l1->val<=l2->val) {
                //直接指向l1,就不用再建立新的ListNode了.
                cur->next = l1;
                l1 = l1->next;
            }
            else {
                cur->next = l2;
                l2 = l2->next;
            }
            cur = cur->next;
        }
        if(l1==NULL) cur->next = l2;
        if(l2==NULL) cur->next = l1;
        return dummy->next;
    }
};