[LeetCode] 330. Patching Array 数组补丁
Given a sorted positive integer array nums and an integer n, add/patch elements to the array such that any number in range [1, n]
inclusive can be formed by the sum of some elements in the array. Return the minimum number of patches required.
Example 1:
nums = [1, 3]
, n = 6
Return 1
.
Combinations of nums are [1], [3], [1,3]
, which form possible sums of: 1, 3, 4
.
Now if we add/patch 2
to nums, the combinations are: [1], [2], [3], [1,3], [2,3], [1,2,3]
.
Possible sums are 1, 2, 3, 4, 5, 6
, which now covers the range [1, 6]
.
So we only need 1
patch.
Example 2:
nums = [1, 5, 10]
, n = 20
Return 2
.
The two patches can be [2, 4]
.
Example 3:
nums = [1, 2, 2]
, n = 5
Return 0
.
Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.
Java:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 | public int minPatches( int [] nums, int n) { long miss = 1 ; int count = 0 ; int i = 0 ; while (miss <= n){ if (i<nums.length && nums[i] <= miss){ miss = miss + nums[i]; i++; } else { miss += miss; count++; } } return count; } |
Python:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 | class Solution( object ): def minPatches( self , nums, n): """ :type nums: List[int] :type n: int :rtype: int """ patch, miss, i = 0 , 1 , 0 while miss < = n: if i < len (nums) and nums[i] < = miss: miss + = nums[i] i + = 1 else : miss + = miss patch + = 1 return patch |
C++:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 | class Solution { public : int minPatches(vector< int >& nums, int n) { long miss = 1, res = 0, i = 0; while (miss <= n) { if (i < nums.size() && nums[i] <= miss) { miss += nums[i++]; } else { miss += miss; ++res; } } return res; } }; |
C++:
1 2 3 4 5 6 7 8 9 10 11 12 13 | class Solution { public : int minPatches(vector< int >& nums, int n) { long miss = 1, k = nums.size(), i = 0; while (miss <= n) { if (i >= nums.size() || nums[i] > miss) { nums.insert(nums.begin() + i, miss); } miss += nums[i++]; } return nums.size() - k; } }; |
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