Split Array Largest Sum——Difficulty:Hard
来源:互联网 发布:看莆田电视台软件 编辑:程序博客网 时间:2024/05/09 07:32
Problem :
Given an array which consists of non-negative integers and an integer m, you can split the array into m non-empty continuous subarrays. Write an algorithm to minimize the largest sum among these m subarrays.
Note:
Given m satisfies the following constraint: 1 ≤ m ≤ length(nums) ≤ 14,000.
Example:
Input:nums = [7,2,5,10,8]m = 2Output:18Explanation:There are four ways to split nums into two subarrays.The best way is to split it into [7,2,5] and [10,8],where the largest sum among the two subarrays is only 18.
**
Algorithm:
**
在网上查到一种很巧妙地方法,用的二分查找。题目把一个数组分成连续的m个数组,得到这些数组求和后的最大值,我们要找的就是这个最大值中的最小值。
很容易知道,这个最大值的范围肯定是在
(原数组的最大值,原数组求和)这个范围内的。
所以就以这两个值为left和right进行二分查找,每次查找都是看对于当前这个mid值作为数组的最大值中的最小值,这个数组是否能满足分成m份的要求,如果能,就说明我们要得到的值还在mid的左边,如果不行,说明在mid右边。
时间复杂度:二分插值O( logn)*判断是否满足条件O(n)=O(nlogn)
**
Code:
class Solution {public: int splitArray(vector<int>& nums, int m) { long long int left=0; long long int right=0; for(int i=0;i<nums.size();i++) { if(nums[i]>left) left=nums[i]; right+=nums[i]; } while(right>left) { long long int mid=(right+left)/2; if(check(nums,m,mid)) { right=mid; } else left=mid+1; } return left; } bool check(vector<int>& nums,int m,int mid) { int c=1; int sum=0; for(int i=0;i<nums.size();i++) { sum+=nums[i]; if(sum>mid) { sum=nums[i]; c++; if(c>m) return false; } } return true; }};
0 0
- Split Array Largest Sum——Difficulty:Hard
- leetcode 410.Split Array Largest Sum(Hard)
- (算法分析Week14)Split Array Largest Sum[Hard]
- Split Array Largest Sum
- Split Array Largest Sum
- Split Array Largest Sum
- 算法练习(11) —— Split Array Largest Sum
- 算法练习(11) —— Split Array Largest Sum
- 算法练习(11) —— Split Array Largest Sum
- LeetCode: Split Array Largest Sum
- 410. Split Array Largest Sum
- 410. Split Array Largest Sum
- 410. Split Array Largest Sum
- [Leetcode] Split Array Largest Sum
- 410. Split Array Largest Sum
- 410. Split Array Largest Sum
- 410. Split Array Largest Sum
- 410. Split Array Largest Sum
- Deep-Learning NotePad2 : Deep Neural network
- [luogu1843] 奶牛晒衣服
- jQuery 之 each 函数 break 和 continue 功能
- HYSBZ 2456: mode
- 十进制数转化其他进制
- Split Array Largest Sum——Difficulty:Hard
- uva 101 木块问题 The Blocks Problem
- linux/unix下 pid文件作用浅析
- io
- Android代码混淆
- 10月17号
- Android弹出的对话框显示输入框
- python IDE安装
- OC的NSLog输出格式