最近需要面试技术人员，接触了codility，顺便给自己来个自我检验。
题目：
Write a function:
int solution(vector<int> &A);
that, given an array A of N integers, returns the smallest positive integer (greater than 0) that does not occur in A.
For example, given A = [1, 3, 6, 4, 1, 2], the function should return 5.
Given A = [1, 2, 3], the function should return 4.
Given A = [−1, −3], the function should return 1.
Assume that:
- N is an integer within the range [1..100,000];
- each element of array A is an integer within the range [−1,000,000..1,000,000].
Complexity:
- expected worst-case time complexity is O(N);
- expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).
解题要点：
- 经过分析，结果范围一定是在[1, N+1]区间之内；//分析出这一点对于解题很重要，我最开始没想到这一点，进行了一些多余操作
- 一定注意对时间复杂度的控制，本题目难度较低，算法的优化主要是在时间复杂度方面；
参考答案：
由于时间比较紧，写的不规范的地方请看官们多担待
int solution(vector<int> &A) {
// write your code in C++14 (g++ 6.2.0)
vector<int> temp(A.size());
for (unsigned int i=0; i<A.size(); i++)
{
int val = A[i];
if (val <= 0 || val > A.size()) continue;
temp[val-1] = 1;
}
unsigned int k=0;
for (;k<temp.size();k++)
{
if (temp[k]==0) break;
}
return k+1;
}
转载请注明原作者及原链接
