LeetCode41/19 First Missing Positive/Remove Nth Node From End of List ****

素颜马尾好姑娘i 2022-08-07 13:38 9阅读 0赞

一：leetcode41 [First Missing Positive][]

题目：

Given an unsorted integer array, find the first missing positive integer.

For example,  
Given `[1,2,0]` return `3`,  
and `[3,4,-1,1]` return `2`.

Your algorithm should run in O(n) time and uses constant space.

分析：此题关键O(n)和常量空间的限制。常量空间的话第一可以考虑是不是固定数量的几个变量能搞定；否则可以考虑是不是问题本身已经提供了足够的空间。当然这里O(1)就可以解决。

class Solution {
    public:
        int firstMissingPositive(int A[], int n) {
            for(int i = 0; i < n; i++){
                while(A[i] > 0 && A[i] <= n &&  A[A[i]-1] != A[i]){  //  比如1 2 5 6 7    此时的A[2]的5就需要与A[4]交换
                    swap(A[i], A[A[i]-1]); 注意条件是让别人正确，不是自己位置A[i]!=i+1 否则如[1 1]就会陷入死循环
                }
            }
            for(int i = 0; i < n; i++)
                if(A[i] != i+1) return i+1;
            return n+1;
            
        }
    };

很多人认为这不是O(N)时间，其实虽然当条件满足，i没有++，但是此时可以是A\[A\[i\]-1\]的值变成了A\[i\]，当循环到下标A\[i\]-1，条件肯定不满足，故还是O(N)时间。当然了本问题没有必要限制数据不重复。。。It is amazing!!

二：LeetCode19 [Remove Nth Node From End of List][]

题目：

Given a linked list, remove the nth node from the end of list and return its head.

For example,

Given linked list: 1->2->3->4->5, and n = 2.
    
       After removing the second node from the end, the linked list becomes 1->2->3->5.

分析：这题给出两种方法，方法一需要额外的空间复杂度O(N),,方法二是双指针的思想，关键在于很难想到啊

/**
     * Definition for singly-linked list.
     * struct ListNode {
     *     int val;
     *     ListNode *next;
     *     ListNode(int x) : val(x), next(NULL) {}
     * };
     */
    /*class Solution {
    public:
        ListNode *removeNthFromEnd(ListNode *head, int n) {    // 需要空间复杂度O(N)
            ListNode *p = head;
            while(p != NULL){
                vec.push_back(p);
                p = p->next;
            }
            int len = vec.size();
            if(n == len) head = head->next;
            else vec[len-n-1]->next = vec[len-n]->next;
            return head;
        }
    private:
        vector<ListNode*> vec;
    };*/
    
    class Solution {
    public:
        ListNode *removeNthFromEnd(ListNode *head, int n) {
            ListNode *front= head, *behind=head;
            while(front != NULL){
                front = front->next;
                if(n-- < 0) behind = behind->next;
            }
            if(n >= 0)head = head->next;
            else behind->next = behind->next->next;
            return head;
        }
    
    };

[First Missing Positive]: https://leetcode.com/problems/first-missing-positive/
[Remove Nth Node From End of List]: https://leetcode.com/problems/remove-nth-node-from-end-of-list/