41. First Missing Positive (sort) O(n) time

Given an unsorted integer array, find the smallest missing positive integer.

Example 1:

Input: [1,2,0]
Output: 3

Example 2:

Input: [3,4,-1,1]
Output: 2

Example 3:

Input: [7,8,9,11,12]
Output: 1

Note:

Your algorithm should run in O(n) time and uses constant extra space.

solution: compare the element with index + 1, if not equal, swap them.

class Solution {
    public int firstMissingPositive(int[] nums) {
        //preprocess the string
        //deal with duplicate elements--two pointers
        //sorted to remove 
        
        for(int i = 0; i<nums.length; i++){
            //check nums[i] != i+1
            while(nums[i] > 0 && nums[i] <= nums.length && nums[i] != nums[nums[i] -1]){
                //swap them
                /*int temp = nums[i];
                nums[i] = nums[nums[i] - 1];
                nums[temp - 1] = temp;*/
                int temp = nums[nums[i] - 1];
                if(temp == nums[i]) break; //check the duplicate elemnt
                nums[nums[i] - 1] = nums[i];
                nums[i] = temp;
            }
        }
        for(int i = 0; i<nums.length; i++){
            if(nums[i] <= 0 || nums[i] != i+1) 
                return i+1;
        }
        return nums.length + 1;
    }
}

Easily, to solve this problem, we can use an extra space to solve this, like a hash set or array.

good reference: https://blog.csdn.net/SunnyYoona/article/details/42683405