描述
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.
样例
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.
dalao思路： https://www.cnblogs.com/grandyang/p/5165821.html
另外做这道题的时候没想通一个问题，如果n个数的和为m，那么这n个数的组合可以构成1-m的任何数
public int minPatches2(int[] nums, int n) {//long miss = 1;int res = 0;int i = 0;while (miss <= n) {if (i < nums.length && nums[i] <= miss) {miss += nums[i++];} else {miss += miss;++res;}}return res;
}summary：用一个变量表示范围内最符合/不符合条件的数，并以此作为边界，一步步处理