给定一个未排序的整数数组，找出其中没有出现的最小的正整数。
示例 1:
输入: [1,2,0] 输出: 3示例 2:
输入: [3,4,-1,1] 输出: 2示例 3:
输入: [7,8,9,11,12] 输出: 1说明:
你的算法的时间复杂度应为O(n)，并且只能使用常数级别的空间。
Put each number in its right place.For example：When we find 5, then swap it with A[4]. At last, the first place where its number is not right, return the place + 1. class Solution { public: int firstMissingPositive(vector<int>& nums) { int n = nums.size(); for (int i = 0; i < n; ++i) { while (nums[i] > 0 && nums[i] <= n && nums[nums[i] - 1] != nums[i]) { swap(nums[i], nums[nums[i] - 1]); } } for (int i = 0; i < n; ++i) { if (nums[i] != i + 1) return i + 1; } return n + 1; } };参考资料
1. My short c++ solution, O(1) space, and O(n) time
