First Missing Positive

    题意是给一个未排序的数组，可能包含正数、负数、0。求第一个未出现的正数，例如[1, 2, 0]，由于第一个未出现的正数是3，返回3；[3, 4, -1, 1]第一个未出现的正数是2，返回2。算法要求O(n)时间复杂度及常数空间复杂度。

// Stable unguarded_partition, always put t to the left side.
// If t is larger than [begin, end), return end.
// If t is smaller than [begin, end), return begin.
template<typename T>
T *unguarded_partition(T *begin, T *end, T t)
{
    while (true)
    {
        while (*begin <= t && begin < end) ++begin;
        --end;
        while (*end > t && begin < end) --end;

        if (!(begin < end)) return begin;

        iter_swap(begin, end);
        ++begin;
    }
}

int firstMissingPositive(int A[], int n)
{
    if (A == NULL || n <= 0) return 1;

    // positive[0] is 0 or the last negative number or invalid position.
    int *positive = unguarded_partition(A, A + n, 0) - 1;
    int positive_len = A + n - positive;

    // Make sure not access positive[0].
    for (int i = 1; i < positive_len; ++i)
    {
        int val = abs(positive[i]);

        // Flip positive[val] to negative.
        if (val > 0 && val < positive_len && positive[val] > 0)
        {
            positive[val] = -positive[val];
        }
    }

    // The first positive index is the first missing positive number.
    for (int i = 1; i < positive_len; ++i)
    {
        if (positive[i] > 0)
        {
            return i;
        }
    }

    // Or the first missing positive num is not in [1, positive)
    return positive_len;
}