Very quick Java solution

  • 0
    import java.util.*;
    public class Solution {
        public int firstMissingPositive(int[] nums) {
            int n = nums.length;
            // 1 is the first positive integer not in an empty list
            if (n == 0)
                return 1;
            // We use a hash set to reduce the problem of determining if an integer is in nums to O(1) time
            HashSet<Integer> numsSet = new HashSet<Integer>();
            for (int i : nums) {
            // There can be no more than n integers in nums, so iterate through the first integers,
            // returning the first one not found in nums
            for (int potential = 1; potential <= n; potential++) {
                if (!numsSet.contains(potential))
                    return potential;
            // The first n integers are in there, so n+1 must be the first missing positive integer
            return n+1;

  • 0

    constant space requirement?

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