1. 题目描述
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.
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.
2. 思路
- 用max_bound当前数组中的数可以表示[1, max_bound)
- 当遍历数组下一个数时
- 如果num[i] <= max_bound, 那么num[i] 可以由前面的数字表示出来， 即不会有断层，结合这个新的数，我们可以表示连续的[1, max_bound + num[i])
- 如果 num[i] > max_bound, 那么[max_bound, num[i])之间的数字就无法表示出来了， 所以我们需要插入一个数字，来防止这种不连续性，当我们添加的数字为max_bound时，可以保证新的覆盖范围最大，即为[1, max_bound * 2)
- 遍历完整个数组或者max_bound > n 时停止。
3. 代码
class Solution {
public:
int minPatches(vector<int>& nums, int n) {
int cnt = 0, i = 0;
long max_bound = 1;
while (max_bound <= n) {
if (i < nums.size() && max_bound >= nums[i]) {
max_bound += nums[i++];
} else {
max_bound *= 2;
++cnt;
}
}
return cnt;
}
};