330. Patching Array ！！！！！！！！

Q:

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.

A:

Sol 1 ： 挨个找，如果找不到，就加入。但是这样不能保证最优。
Sol 2:  保持found list，和non-found list， 找出加入一个数字，可以保证满足最多的（可是这样
但是这样，一是麻烦，而是不保证正确，三是LTE
Sol 3:

public class Solution {
    public int minPatches(int[] nums, int n) {
        int count = 0, i = 0;
        long findMaxRight = 1;// this maxRight has not been found yet
        while(findMaxRight <= n){
            if( i<nums.length && nums[i] <= findMaxRight ){
                findMaxRight += nums[i++];
            }else{
                findMaxRight = 2*findMaxRight;
                count++;
            }
        }
        return count;
    }
}

Mistakes:
1: findMaxRight is exclusive.   thus its initial value can be set to 1.    WTF
2:  use long , instead of int , to avoid overflow.