330. Patching Array

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.

解题思路:每次补的数都是从1->n中第一个到不了的数，当前能到的最远的数为x，则下一个补的数一定是x+1，这样才能使得他到的位置最远。另外就是已知的数的利用比较不容易想到。

class Solution {
public:
    int minPatches(vector<int>& nums, int n) {
       long long int mmax=1,count=0,i=0;
       while(mmax<=n){
           if(i<nums.size()&&nums[i]<=mmax)mmax+=nums[i++];
           else mmax+=mmax,count++;
       }
       return count;
    }
};