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;
     }
 }