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.

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思路：我们依次将数组中缺少的数字加入数组nums.设置当前数字组合相加的范围是[1,total),total初始设置为1.当nums[i]存在且nums[i]<=total时，相加的范围可以拓展到[1,total+nums[i]).nums[i]最大可以等于total，否则我们利用贪心的策略，尽可能快地提高total的值，向数组nums中插入total。重复上述循环直至total>n.最终增加的数的个数就是数组nums大小的变化。

class Solution {

public:

    int minPatches(vector<int>& nums, int n) {

        int N =nums.size();

        long total =;

        int i=;

        while(total<=n){

            if(i<nums.size() &&nums[i]<=total){

                total+=nums[i++];

            }

            else{

                nums.insert(nums.begin()+i,total);

            }

        }

        return nums.size()-N;

    }

};