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<int>& 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<nums.length && nums[i]<=miss){
miss = miss + nums[i];
i++;
}else{
added++;
miss += miss;
}
}
return added;
}
}
