标签：编译器 des 比赛 ace img system stream cin png
题目描述
Description
Given nn non-negative integers, please find the least non-negative integer that doesn’t occur in the nn numbers.
Input
The first line is an integer T, representing the number of test cases.(T≤10)
The first line of each test case is an integer n (n≤2?10?^5??).
The second line of each test case are nn non-negative
integers a?i?? (a?i??<2?^31??).
Output
For each test case, output a line with the answer.
输入样例
2
4
4 0 1 3
2
1 1
输出样例
2
0
分析：暴力数字范围给定，暴力搜索其实也不是不可以，但是肯定会超时，网上有很多解法针对的都是数组元素不重复的情况，参考价值也不是很大，后来我自己想了一种方法，需要先进行排序，之后从头遍历，需要记录下来重复数字的重复次数，根据数组下标和重复次数之间的 关系，得到解答。时间复杂度O(n)，空间复杂度O(1)
代码：
1 #include<algorithm> 2 #include<iostream> 3 #include<string> 4 #include<vector> 5 using namespace std; 6 int main() 7 { 8 int n; 9 cin >> n; 10 // 打比赛时使用GCC编译器，务必要将两个>之间打上空格，以免误判 11 vector<vector<int> > v(1001,vector<int>(200001,INT_MAX)); 12 vector<int> len(1001); 13 for (int i = 0; i < n; i++) 14 { 15 //int m; 16 cin >> len[i]; 17 for (int j = 0; j < len[i]; j++) 18 cin >> v[i][j]; 19 } 20 for (int i = 0; i < n; i++) 21 { 22 int res = 0; 23 int repNum = 0; 24 sort(v[i].begin(), v[i].end()); 25 if (v[i][0] > 0) 26 { 27 cout << 0 << endl; 28 continue; 29 } 30 else 31 { 32 int j = 1; 33 for ( ; j < len[i]; j++) 34 { 35 if (v[i][j] == v[i][j - 1]) 36 { 37 repNum++; 38 continue; 39 } 40 if (v[i][j] != j - repNum) 41 { 42 cout << j - repNum << endl; 43 break; 44 } 45 } 46 if (j == len[i]) 47 cout << len[i] << endl; 48 } 49 } 50 system("pause"); 51 return 0;
标签：编译器 des 比赛 ace img system stream cin png
原文地址：https://www.cnblogs.com/dapeng-bupt/p/8746792.html
