SpringMVC實現上傳和下載
阿新 • • 發佈:2020-08-12
問題:
給定一個升序陣列,求第k個該陣列缺失的數。
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
解法:
解法一:二分查詢(Binary Search)
找到arr[m]之前缺失數字個數arr[m]-(m+1),若>=k,
那要找的缺失數為:已有的數(未缺失)+缺失的第k個=m+k
程式碼參考:
1 class Solution { 2 public: 3 int findKthPositive(vector<int>& arr, int k) { 4 int l = 0, r = arr.size(); 5 while(l<r) { 6 int m = l+(r-l)/2; 7 int count = arr[m]-m-1; 8 if(count>=k) { 9 r = m; 10 } else { 11 l = m+1; 12 } 13 } 14 return l+k; 15 } 16 };
解法二:hash
將已有的數字,存入unordered_set中,
從1開始輪詢每個數字,若沒有在已有的數字列表中,
k--,說明該數字是漏掉的數字。漏掉數字總數-1。
直到k=0,當前數字則是要找的漏掉的第k個數字。返回。
♻️ 優化,在輪詢到已有數字列表的最後一個時,k還沒減完,
則直接返回最後一個數字+當前剩下的k。
程式碼參考:
1 class Solution { 2 public: 3 int findKthPositive(vector<int>& arr, int k) { 4 int res=0; 5 unordered_set<int> arrset(arr.begin(), arr.end()); 6 for(int i=1; i<=arr.back(); i++) { 7 if(arrset.count(i)==0) k--; 8 if(k==0) return i; 9 } 10 return arr.back()+k; 11 } 12 };