[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:

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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:

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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++:

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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++:

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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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