LeetCode No330. Patching Array
2016-06-10 14:02
381 查看
1. 题目描述
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.
2. 思路
用max_bound当前数组中的数可以表示[1, max_bound)当遍历数组下一个数时
如果num[i] <= max_bound, 那么num[i] 可以由前面的数字表示出来, 即不会有断层,结合这个新的数,我们可以表示连续的[1, max_bound + num[i])
如果 num[i] > max_bound, 那么[max_bound, num[i])之间的数字就无法表示出来了, 所以我们需要插入一个数字,来防止这种不连续性,当我们添加的数字为max_bound时,可以保证新的覆盖范围最大,即为[1, max_bound * 2)
遍历完整个数组或者max_bound > n 时停止。
3. 代码
class Solution { public: int minPatches(vector<int>& nums, int n) { int cnt = 0, i = 0; long max_bound = 1; while (max_bound <= n) { if (i < nums.size() && max_bound >= nums[i]) { max_bound += nums[i++]; } else { max_bound *= 2; ++cnt; } } return cnt; } };
4. 参考资料
相关文章推荐
- 详解Android应用中屏幕尺寸的获取及dp和px值的转换
- 基于Android中dp和px之间进行转换的实现代码
- Android中dip、dp、sp、pt和px的区别详解
- LFC1.0.0 版本发布
- leetcode 179 Largest Number
- leetcode 24 Swap Nodes in Pairs
- leetcode 2 Add Two Numbers 方法1
- leetcode 2 Add Two Numbers 方法2
- Android dpi,dip,dp的概念以及屏幕适配
- Android px、dp、sp之间相互转换
- HP data protector软件学习1--基本角色与基本工作流程
- HP data protector软件学习2--软件组成与界面介绍
- leetcode----Longest Substring Without Repeating Characters
- [LeetCode]47 Permutations II
- [LeetCode]65 Valid Number
- [LeetCode]123 Best Time to Buy and Sell Stock III
- [LeetCode] String Reorder Distance Apart
- [LeetCode] Sliding Window Maximum
- [LeetCode] Find the k-th Smallest Element in the Union of Two Sorted Arrays
- [LeetCode] Determine If Two Rectangles Overlap