[leetcode] 1539. Kth Missing Positive Number

Description

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

Find 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

分析

题目的意思是：给你一个递增数组，找出第K个缺省的值。很直接，暴力破解就能够解决，如代码一，找到第K个就行了。我在看别人的解法的时候发现了一个更好的解法，只需要遍历arr数组就行了，每次遍历的时候计算两数之间缺省的值的个数，然后推算出第K个位置，很巧妙哈哈哈。

代码一

class Solution:
    def findKthPositive(self, arr: List[int], k: int) -> int:
        i=1
        j=0
        cnt=0
        n=len(arr)
        res=[]
        while(cnt<k):
            if(j<n and i<arr[j]):
                res.append(i)
                i+=1
                cnt+=1
            elif(j<n and i==arr[j]):
                j+=1
                i+=1
            elif(j>=n):
                res.append(i)
                i+=1
                cnt+=1
        return res[-1]

代码二

class Solution:
    def findKthPositive(self, arr: List[int], k: int) -> int:
        missing=0
        n=len(arr)
        for i in range(n):
            if(i==0):
                missing=arr[i]-1
            else:
                missing+=arr[i]-arr[i-1]-1
            if(missing>=k):
                return arr[i]-(missing-k)-1
        return arr[-1]+k-missing
            
        return res[-1]

参考文献

​​[LeetCode] Python, single array pass. Time: O(N), Space: O(1)​​