[LeetCode] First Missing Positive

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.

思路：新建一个数组B，然后用A[i]的值作为B的index，即B[A[i]] =1，然后遍历一遍B,寻找B[i]！=1的index，不过貌似我的时间复杂度是O(n)，空间也是O（n）啊。。

class Solution {
public:
int firstMissingPositive(int A[], int n)
{
vector<int> B;
B.resize(n + 1);

//cout << "n\t" << n << endl;
//printArray(A, n);
for(int i = 0; i < n; i++)
{
if(A[i] <= n && A[i] >0)
{
// do A[i]'s value as B's index
B[A[i]] = 1;
}
}
//printVector(B);

for(size_t i = 1; i < B.size(); i++)
{
if(B[i] != 1)
return i;
}
return  n+1;
}
};

为了实现O(n)的空间复杂度，只能使用数组A自身的空间，A[i]保存i+1,不停的交换，具体看注释吧，我觉得写的挺清楚的。。

class Solution {
public:
#if 0
int firstMissingPositive(int A[], int n)
{
vector<int> B;
B.resize(n + 1);

//cout << "n\t" << n << endl;
//printArray(A, n);
for(int i = 0; i < n; i++)
{
if(A[i] <= n && A[i] >0)
{
// do A[i]'s value as B's index
B[A[i]] = 1;
}
}
//printVector(B);

for(size_t i = 1; i < B.size(); i++)
{
if(B[i] != 1)
return i;
}
return  n+1;
}
#endif
int firstMissingPositive(int A[], int n)
{
int i = 0;
int tmp;
// A[i] should store (i + 1)
// it means A[0] stores 1, A[1] stores 2, ...

while(i < n)
{
// A[0~n-1] only can store 1~n;
// A[i] should store i + 1
// A[i] != i +1 menas A[i] should be move to others, but A[i] should be moved where?
// A[i] should be move to A[i]-1
// i.e.: {3, 1, 4, -1}
// A[0]= 3, != (0+1), so 3 shoule be moved. moved to A[0] -1 =2, then check if A[2] and A[0] are equal.
// so swap 3 and 4, then the array is {4, 1, 3 , -1}
// but 4 also doesn't appear at A[0], so go on to swap, until the condition doens't meet, then i++
if(A[i] <= n && A[i] >0 && A[i] != (i+1) && A[A[i] - 1] != A[i] )
{
tmp = A[i];
A[i] = A[tmp - 1];
A[tmp - 1] = tmp;
}
else
i++;
}
//printArray(A, n);

for(int i = 0; i < n; i++)
{
if(A[i] != (i+1) )
return i + 1;
}
return n+1;

}
};