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.

思路：设置一个标志位miss，用于记录可以到达的范围（0，miss），类似于数学中的开区间,初始化为1.  遍历数组中的元素，当元素大于miss的时候，说明出现了不连续的情况，需要加上miss；如果元素小于miss,说明元素可达，则加上该元素扩大区间。

int minPatches(vector<int>& nums, int n) {
    long miss = 1, added = 0, i = 0;
    while (miss <= n) {
        if (i < nums.size() && nums[i] <= miss) {
            miss += nums[i++];                         //情况一：miss < num[i],不需要增加miss元素
        } else {
            miss += miss;                              //情况二：miss > num[i],需要增加miss元素
            added++;
        }
    }
    return added;
}