Given an unsorted integer array, find the first missing positive integer.

For example,
Given [1,2,0] return 3,
and [3,4,-1,1] return 2.

Your algorithm should run in O(n) time and uses constant space.

int firstMissingPositive(vector<int>& nums) {
int n = nums.size(), t, i;
for(i = ; i < n; i++)
{
t = nums[i];
while(t > && t < n && nums[t-] != t)
{
swap(nums[i], nums[t-]);
t = nums[i];
}
}
for(i = ; i < n; i++)
{
if(nums[i] != i+)
return i+;
}
return n+;
}

Idea:
*
* We can move the num to the place whcih the index is the num.
*
* for example, (considering the array is zero-based.
* 1 => A[0], 2 => A[1], 3=>A[2]
*
* Then, we can go through the array check the i+1 == A[i], if not ,just return i+1;

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