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：因为要求不能用到额外的空间，题目有暗示：答案在[1,n+1]之间。每个位置都试着将这个位置的值换到对应的下标，这样第一个位置出现不是相应的值的时候就是答案，还有就是要是都能对应，那么n+1就是解

class Solution {
public:
	int firstMissingPositive(int A[], int n) {
		for (int i = 0; i < n; i++) A[i]--;

		for (int i = 0; i < n; i++) {
			while (A[i] != i && A[i] >= 0 && A[i] < n) {
				if (A[i] == A[A[i]]) break;
				swap(A[i], A[A[i]]);
			}
		}

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

};