描述

Given N integers, you are supposed to find the smallest positive integer that is NOT in the given list.

指定输入

Each input file contains one test case. For each case, the first line gives a positive integer N (≤10^5). Then N integers are given in the next line, separated by spaces. All the numbers are in the range of int.

指定输出

Print in a line the smallest positive integer that is missing from the input list.

输入样例

10
5 -25 9 6 1 3 4 2 5 17

输出样例

7

分析

其实就是找出连续最小的正数,网上有一种解法,怎么说呢,想法很清奇但也很有效,就是下面这个

#include <bits/stdc++.h>
using namespace std;
const int maxn = 100000 + 10;
int n;
int vis[maxn];
int main() {
    scanf("%d", &n);
    int MAX = -1;
    for (int i = 0; i < n; i++) {
        int a;
        scanf("%d", &a);
        if (a > 0 && a < maxn) vis[a] = 1;//注意这里
    }
    for (int i = 1; i < maxn; i++) {
        if (vis[i] == 0) {
            printf("%d\n", i);
            break;
        }
    }
 
    return 0;
}

但python很难做到,因为python只有列表(list)类型,很难初始化一个这样庞大的数组(又或许是我队python掌握得不够深),我的想法是下面(本来使用python写的,但一不小心被我删了),只有c++版本,总是有一个错误但是没有搞出来,很气

#include <iostream>
#include <algorithm>

using namespace std;

int main()
{
	int n;
	cin >> n;
	int in[11000];
	for (int i = 0; i < n; i++) {
		cin >> in[i];
	}
	sort(in, in + n);
	int left_p = 0;
	int right_p = 1;

	for (int i = 0; i < n-1; i++) {
		if (in[left_p] < 0) {
			left_p += 1;
			right_p += 1;
			if (right_p == n) {
				if (in[n - 1] <= 0) {
					cout << 1;
				}
				else
				{
					cout << in[n - 1] + 1;
				}
			}
		}
		else if (in[right_p] - in[left_p] == 1 || in[right_p] - in[left_p] == 0) {
			left_p += 1;
			right_p += 1;
			if (right_p == n) {
				cout << in[left_p] + 1;
				break;
			}
		}
		else
		{
			cout << in[left_p] + 1;
			break;
		}
	}
}
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