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.
本题基于在一个数列中，
1+A1+A2+......+Ak >= Ak+1，则[1,k]连续
public class Solution {
public int minPatches(int[] nums, int n) {
int miss = 1;
int add = 0;
int i = 0;
while(miss <= n) {
//两个条件<br>1. i < nums.length, 数组的全部累加和小于n<br>2. miss >= nums[i], 这条就是上面提到的数学原理
if(i < nums.length && miss >= nums[i]) {
miss = miss + nums[i];
i++;
}
//如果累加的和小于nums[i+1],则补上这个sum（当条件不满足时，补上1+A1+A2+...+Ak)
else {
add++;
//第一个miss是sum，第二个miss是要插入数组的值,新的miss是添加完这个数的sum
miss = miss + miss;
}
}
return add;
}
}
需要注意的是，这段Java代码在n取到极值232-1时，会有溢出，需要将int转换为long