Description Link to heading

41. First Missing Positive (Hard)

Given an unsorted integer array nums, return the smallest missing positive integer.

You must implement an algorithm that runs in O(n) time and uses constant extra space.

Example 1:

Input: nums = [1,2,0]
Output: 3
Explanation: The numbers in the range [1,2] are all in the array.

Example 2:

Input: nums = [3,4,-1,1]
Output: 2
Explanation: 1 is in the array but 2 is missing.

Example 3:

Input: nums = [7,8,9,11,12]
Output: 1
Explanation: The smallest positive integer 1 is missing.

Constraints:

  • 1 <= nums.length <= 10⁵
  • -2³¹ <= nums[i] <= 2³¹ - 1

Solution Link to heading

We can mark nums[i]，we set the negative elements in nums as n + 1, and num = abs(nums[i]), ans set nums[num - 1] = -nums[num - 1].

Or we can swap.

Code Link to heading 

class Solution {
public:
    int firstMissingPositive(vector<int>& nums) {
        int n = nums.size();
        for (int& num: nums) {
            if (num <= 0) {
                num = n + 1;
            }
        }
        for (int i = 0; i < n; ++i) {
            int num = abs(nums[i]);
            if (num <= n) {
                nums[num - 1] = -abs(nums[num - 1]);
            }
        }
        for (int i = 0; i < n; ++i) {
            if (nums[i] > 0) {
                return i + 1;
            }
        }
        return n + 1;
    }
};
// @lc code=end