Last Updated on 2023-03-06 by Clay
題目
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 <= 10001 <= arr[i] <= 10001 <= k <= 1000arr[i] < arr[j]for1 <= i < j <= arr.length
Follow up:
Could you solve this problem in less than O(n) complexity?
題目指定我們要找到第 k 個不存在於陣列 arr 的正整數，並將其回傳。要注意的是，第 k 個正整數的值可能會遠超過陣列 arr 的範圍。
解題思路
這題不愧是 Easy 題，直接從 1 開始計數就能獲得不錯的跑分結果。
簡單來說，就是我們設定一個 missing 的變數，一開始初始化為 1；接著我們逐項與 arr 中的值做比較。
- 如果與
arr中的值一致，則將missing加 1 - 如果與
arr中的值不一致，則除了將missing加 1 外，也要把k減去 1；一但k減至 0，則將當前的missing回傳，因為它就是第 k 個遺失的正整數
值得注意的是，可能整個 arr 走完都還沒找到第 k 個正整數，這時候只要透過 k 剩下的值，我們便能直接計算出最後要的正整數。
複雜度
| Time Complexity | O(n) |
| Space Complexity | O(1) |
C++ 範例程式碼
class Solution {
public:
int findKthPositive(vector<int>& arr, int k) {
// Init
int missing = 1;
// Comparing with the `arr` one by one
for (int i=0; i<arr.size(); ++i) {
// If `missing` not in `arr`, that means we can count down `k` value
while (missing != arr[i]) {
--k;
if (k == 0) {
return missing;
}
++missing;
}
++missing;
}
// If `arr` is over and we do not find the answer, we can calculate the results directly
return missing + k - 1;
}
};Python 範例程式碼
class Solution:
def findKthPositive(self, arr: List[int], k: int) -> int:
# Init
missing = 1
# Comparing with the `arr` one by one
for num in arr:
# If `missing` not in `arr`, that means we can count down `k` value
while missing != num:
if k == 1:
return missing
k -= 1
missing += 1
missing += 1
# If `arr` is over and we do not find the answer, we can calculate the results directly
return missing + k - 1