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Leetcode: Patching Array

时间:2016-01-28 07:05:01      阅读:282      评论:0      收藏:0      [点我收藏+]

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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.

reference: https://leetcode.com/discuss/82822/solution-explanation

Let miss be the smallest sum in [1,n] that we might be missing. Meaning we already know we can build all sums in [1,miss). Then if we have a number num <= miss in the given array, we can add it to those smaller sums to build all sums in [1,miss+num). If we don‘t, then we must add such a number to the array, and it‘s best to add miss itself, to maximize the reach.

 

 1 public class Solution {
 2     public int minPatches(int[] nums, int n) {
 3         long missed=1;
 4         int added=0, i=0;
 5         while (missed <= n) {
 6             if (i<nums.length && nums[i]<=missed) {
 7                 missed = missed + nums[i];
 8             }
 9             else {
10                 added++;
11                 missed = missed + missed;
12                 i--;
13             }
14             i++;
15         }
16         return added;
17     }
18 }

 

Leetcode: Patching Array

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原文地址:http://www.cnblogs.com/EdwardLiu/p/5165054.html

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