[LeetCode] 330. Patching Array 数组补丁

2021年09月15日 阅读数:1
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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.html

Example 1:
nums = [1, 3]n = 6
Return 1.java

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.python

Example 2:
nums = [1, 5, 10]n = 20
Return 2.
The two patches can be [2, 4].post

Example 3:
nums = [1, 2, 2]n = 5
Return 0.ui

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.this

Java:url

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:code

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++:orm

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++:htm

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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