[LeetCode] 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.
解题思路
如果数组当前能够表示的范围为[1, total)内的全部数字。那么向数组中加入元素add能够将表示范围扩充至[1, total + add)。当中add≤total。当且仅当add=total时取到范围上限[1, 2 * total)。
- 当数组中有小于等于
add的元素时。则利用数组中的元素。 - 若没有,则加入新元素
add。
实现代码
//Runtime: 1 ms
public class Solution {
public int minPatches(int[] nums, int n) {
long miss = 1;
int add = 0;
int i = 0;
while (miss <= n) {
if (i < nums.length && nums[i] <= miss){
miss += nums[i++];
} else {
miss += miss;
add += 1;
}
}
return add;
}
}

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