[LeetCode] Patching Array

Posted 2020-06-09 楚兴

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

假设数组当前可以表示的范围为[1, total)内的所有数字，那么向数组中添加元素add可以将表示范围扩充至[1, total + add)，其中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，则添加新元素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;
    }
}