First Missing Positive 解答

Question

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.

Solution 1 -- HashSet

Note the problem is to find first missing integer, ie, if input is [4,5,7,8], we should return 1.

Naive way is to traverse from 1 to length, find whether they are in original input set. Time complexity O(n), space cost O(n).

public class Solution {
    public int firstMissingPositive(int[] nums) {
        int length = nums.length;
        Set<Integer> set = new HashSet<Integer>();
        for (int i = 0; i < length; i++)
            set.add(nums[i]);
        int result = 1;
        for (int i = 1; i <= length; i++) {
            if (!set.contains(i)) {
                result = i;
                break;
            } else {
                // Should increase here
                // Consider input [1]
                result++;
            }
        }
        return result;
    }
}

Solution 2

Key is to put i on position (i - 1). Time complexity O(n), space cost O(1).

public class Solution {
    public int firstMissingPositive(int[] nums) {
        int length = nums.length;
        // Put i on (i - 1) position
        // Consider three conditions: 1. nums[i] <= 0 2. nums[i] > length 3. duplicates
        for (int i = 0; i < length; i++) {
            int target = nums[i];
            if (target != i + 1) {
                // nums[i] <= 0 nums[i] > length
                if (target <= 0 || target > length)
                    continue;
                // Duplicate
                if (nums[target - 1] == target) {
                    nums[i] = 0;
                    continue;
                }
                // Swap target with nums[target - 1]
                int tmp = nums[target - 1];
                nums[target - 1] = target;
                nums[i] = tmp;
                // Still need to check nums[i] after swap;
                i--;
            }
        }
        
        int result = length + 1;
        // Check
        for (int i = 0; i < length; i++) {
            if (nums[i] != i + 1) {
                result = i + 1;
                break;
            }
        }
        return result;
    }
}