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.

一刷
题解：
让miss表示[0-n]中最小的一个可能miss掉的sum, 那么我们可以达到[0, miss)所有的sum。
对于当前的num, 如果num<=miss, 那么就可以组成[0, miss+num)的所有sum, 如果不能,则把miss加入其中。

int minPatches(vector& nums, int n) {
    long miss = 1, added = 0, i = 0;
    while (miss <= n) {
        if (i < nums.size() && nums[i] <= miss) {
            miss += nums[i++];
        } else {
            miss += miss;
            added++;
        }
    }
    return added;
}

二刷

class Solution {
    public int minPatches(int[] nums, int n) {
        long miss = 1;
        int added = 0, i=0;
        while(miss<=n){
            if(i