【PAT 甲级】1144 The Missing Number（20 分）

最新推荐文章于 2021-08-18 19:29:53 发布

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本文介绍了一种算法，用于从给定的整数列表中找出最小的未出现的正整数。通过排序输入列表并使用二分查找来提高效率，确保了算法的有效性和准确性。

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题目链接

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

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N ( $\leq 1 0^{ 5 }$ ). Then N integers are given in the next line, separated by spaces. All the numbers are in the range of int.

Output Specification:

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

Sample Input:

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

Sample Output:

代码：

#include <iostream>
#include <sstream>
#include <string>
#include <cstdio>
#include <queue>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int N = 1e5+10;
const ll INF = 0x3f3f3f3f;
const double eps=1e-4;
const int T=3;

int a[N];
int main() {
    int n;
    scanf("%d",&n);
    for(int i=0; i<n; i++) {
        scanf("%d",&a[i]);
    }
    sort(a,a+n);
    int pos=lower_bound(a,a+n,1)-a;
    // printf("A %d\n",a[pos]);
    if(pos>=n)
        printf("1");
    else {
        bool flag=0;
        for(int i=pos,j=1; i<n; i++,j++) {
            if(i>=1&&a[i]==a[i-1]) {
                j--;
                continue;
            }
            if(a[i]!=j) {
                printf("%d",j);
                flag=1;
                break;
            }

        }
        if(!flag) {
            printf("%d",a[n-1]+1);
        }
    }
    return 0;
}