LeetCode:462. Minimum Moves to Equal Array Elements II
2017-09-11 10:49
507 查看
Given a non-empty integer array, find the minimum number of moves required to make all array elements equal, where a move is incrementing a selected element by 1 or decrementing a selected element by 1.
You may assume the array’s length is at most 10,000.
Example:
Input:
[1,2,3]
Output:
2
Explanation:
Only two moves are needed (remember each move increments or decrements one element):
[1,2,3] => [2,2,3] => [2,2,2]
题意:将数组中所有的数转化成相同的值,每次只能对一个数增或减1,求最少的步骤。
分析:这题和编程之美上的电梯停靠问题类似,比那个简单。主要的问题就是确定最终转化的那个数是什么值才能使转化步骤最少。
首先对数据排序。
设最终值为Q,假设Q=最小的数,那么将数组分成两部分,left=0;right=sum(nums)-n*nums[0],那么总的转化步骤count=left+right。然后可以逐渐增加Q:
Q+1: left’=left+1; right’=right-(n-1) ; count’=left’+right’=count+(2-n)
···
Q+nums[1]-nums[0],即Q=nums[1]: left’=left+nums[1]-nums[0]; right’=right-(n-1)(nums[1]-nums[0]; count’=left’+right’=count+(2(nums[1]-nums[0])-n)
···
Q=nums[i]: left’=left+i*(nums[i]-nums[i-1]);
right’=right-(n-i)*(nums[i]-nums[i-1])
4000
count’=left’+right’=count+(2*i-n)*(nums[i]-nums[i-1]); ——(A)
对于通项式(A),当i从小变大的过程中,count先下降在上升,当2*i-n>=0,并且2*(i+1)-n<0时count取最小,即最终转化值Q=中位数时,最终转化值最小。
代码如下:
You may assume the array’s length is at most 10,000.
Example:
Input:
[1,2,3]
Output:
2
Explanation:
Only two moves are needed (remember each move increments or decrements one element):
[1,2,3] => [2,2,3] => [2,2,2]
题意:将数组中所有的数转化成相同的值,每次只能对一个数增或减1,求最少的步骤。
分析:这题和编程之美上的电梯停靠问题类似,比那个简单。主要的问题就是确定最终转化的那个数是什么值才能使转化步骤最少。
首先对数据排序。
设最终值为Q,假设Q=最小的数,那么将数组分成两部分,left=0;right=sum(nums)-n*nums[0],那么总的转化步骤count=left+right。然后可以逐渐增加Q:
Q+1: left’=left+1; right’=right-(n-1) ; count’=left’+right’=count+(2-n)
···
Q+nums[1]-nums[0],即Q=nums[1]: left’=left+nums[1]-nums[0]; right’=right-(n-1)(nums[1]-nums[0]; count’=left’+right’=count+(2(nums[1]-nums[0])-n)
···
Q=nums[i]: left’=left+i*(nums[i]-nums[i-1]);
right’=right-(n-i)*(nums[i]-nums[i-1])
4000
count’=left’+right’=count+(2*i-n)*(nums[i]-nums[i-1]); ——(A)
对于通项式(A),当i从小变大的过程中,count先下降在上升,当2*i-n>=0,并且2*(i+1)-n<0时count取最小,即最终转化值Q=中位数时,最终转化值最小。
代码如下:
class Solution { public int minMoves2(int[] nums) { Arrays.sort(nums); int index=nums.length/2; int count=0; for(int i=0;i<nums.length;i++){ count+=Math.abs(nums[index]-nums[i]); } return count; } }
相关文章推荐
- leetcode 462. Minimum Moves to Equal Array Elements II
- Leetcode-462. Minimum Moves to Equal Array Elements II
- LeetCode - 462. Minimum Moves to Equal Array Elements II - 思路详解 - C++
- leetcode_middle_4_462. Minimum Moves to Equal Array Elements II
- leetcode 462. Minimum Moves to Equal Array Elements II
- 【LeetCode】462. Minimum Moves to Equal Array Elements II 解题报告(Python)
- [leetcode] 462. Minimum Moves to Equal Array Elements II 解题报告
- LeetCode 462. Minimum Moves to Equal Array Elements II
- LeetCode笔记:462. Minimum Moves to Equal Array Elements II
- [LeetCode]462. Minimum Moves to Equal Array Elements II
- LeetCode 462. Minimum Moves to Equal Array Elements II
- Leetcode 462. Minimum Moves to Equal Array Elements II 使数组相等 解题报告
- 【LeetCode】462. Minimum Moves to Equal Array Elements II
- LeetCode "462. Minimum Moves to Equal Array Elements II"
- leetcode_462. Minimum Moves to Equal Array Elements II
- LeetCode | 462. Minimum Moves to Equal Array Elements II 简单数学原理题
- 【LeetCode】 462. Minimum Moves to Equal Array Elements II
- leetcode462. Minimum Moves to Equal Array Elements II
- LeetCode 462. Minimum Moves to Equal Array Elements II
- leetcode 462. Minimum Moves to Equal Array Elements II