330. Patching Array--Avota

问题描述:

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.

题意:

给定非递减正数组nums和正整数n,问最少向数组中添加多少元素使得从nums[ ]中取若干元素的和能够覆盖[1, n]

解题思路:

我们可以试着从1到n检查每个数(记为current)是否满足。对于每一个current,先从输入数组nums中查看是否有满足条件的数(即是否nums[i] <= current,i表示nums数组中的数用到第几个,初始为0),若有则使用并进行i++、current += nums[i](current至current + nums - 1可由current-nums[i]到current - 1分别加上nums[i]得到)操作;若无则添加新元素current并进行current = current * 2操作。
具体的,扫描数组nums,更新原则如下:

  • 若nums[i] <= current , 则把nums[i]用掉(即 i++),同时current更新为current + nums[i];
  • 若nums[i] > current,则添加新的元素current,同时current更新为current * 2.

但须注意:
current从1开始
current可能超过int型,需使用long型

示例代码:

class Solution {
public:
    int minPatches(vector<int>& nums, int n) {
        int len = nums.size();
        if(len == 0){
            return log2(n) + 1;
        }
        
        long current = 1;
        int i = 0, count = 0;
        while(current <= n){
            if(i < len  && nums[i] <= current){
                current += nums[i];
                i++;
            }
            else{
                count++;
                current *= 2;
            }
        }
        return count;
    }
};
posted @ 2016-03-06 22:10  Avota  阅读(193)  评论(0编辑  收藏  举报