leetcode——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.

这道题目一看没有思路，稍微想想，估计每次加入的数字是最小的没有覆盖的数，但是思考到这里并不能AC（超时）。继续思考的话：一个数只能被小于等于本身的数字得到。因此假设[1,x)的数字已经被覆盖，但是还有很多数字没有达到n，那么添加x，既能覆盖x，又能覆盖更多的数，达到[1,2*x)的范围。通过数学归纳可得解法。

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