程式小試身手

解答中的分治法 (Divide-and-conquer algorithm)

class Solution {
    public static ListNode mergeKLists(ListNode[] lists){
    return partion(lists,0,lists.length-1);
}

public static ListNode partion(ListNode[] lists,int s,int e){
    if(s==e)  {return lists[s];}
    if(s<e){
        int q=(s+e)/2; //先帶入0,2得 1,l1=partion(lists,0,1),l1=partion(lists,0,0)得返回,l2=partion(lists,1,1)得返回 進行比較
        ListNode l1=partion(lists,s,q);
        ListNode l2=partion(lists,q+1,e);
        return merge(l1,l2);
    }else
        return null;
}

//This function is from Merge Two Sorted Lists.
public static ListNode merge(ListNode l1,ListNode l2){
    if(l1==null) return l2;
    if(l2==null) return l1;
    if(l1.val<l2.val){
        l1.next=merge(l1.next,l2);
        return l1;
    }else{
        l2.next=merge(l1,l2.next);
        return l2;
    }
}
}