41. First Missing Positive

Given an unsorted integer array nums. Return the smallest positive integer that is not present in nums.

You must implement an algorithm that runs in O(n) time and uses O(1) auxiliary 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

class Solution {
public:
    int firstMissingPositive(vector<int>& nums) {
        int n = nums.size();
        for(int i = 0; i < n; i++) {
            // 若數字大於 0 且小於等於 n，且該位置上的數字不等於當前數字，則進行交換
            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;
    }
};

• T: $O(N)$
• S: $O(1)$