leetcode-1539-easy

Kth Missing Positive Number

Given an array arr of positive integers sorted in a strictly increasing order, and an integer k.

Return the kth positive integer that is missing from this array.

Example 1:

Input: arr = [2,3,4,7,11], k = 5
Output: 9
Explanation: The missing positive integers are [1,5,6,8,9,10,12,13,...]. The 5th missing positive integer is 9.
Example 2:

Input: arr = [1,2,3,4], k = 2
Output: 6
Explanation: The missing positive integers are [5,6,7,...]. The 2nd missing positive integer is 6.
Constraints:

1 <= arr.length <= 1000
1 <= arr[i] <= 1000
1 <= k <= 1000
arr[i] < arr[j] for 1 <= i < j <= arr.length
Follow up:

Could you solve this problem in less than O(n) complexity?

思路一：把 k 想象成一个 [0, k] 范围数轴，只要数组的值出现在数轴里面，k 就往右移动，只要计算 [0, k] 里面出现过多少数字，就可以统计第 kth 大的数字

    public int findKthPositive(int[] arr, int k) {
        for (int i : arr) {
            if (k >= i) k++;
        }

        return k;
    }