LeetCode:453. Minimum Moves to Equal Array Elements

最新推荐文章于 2024-11-09 07:15:00 发布

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最新推荐文章于 2024-11-09 07:15:00 发布

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原文链接：http://www.cnblogs.com/luluqiao/p/6369657.html

本文介绍了解决LeetCode上编号为453的问题——如何用最少的步数使数组元素相等的方法。通过寻找数组中的最小值并计算所有元素与该最小值之间的差值总和来得出答案。

 1 package Today;
 2 //LeetCode:453. Minimum Moves to Equal Array Elements
 3 /*
 4 Given a non-empty integer array of size n, find the minimum number of moves required to make all array 
 5 elements equal, where a move is incrementing n - 1 elements by 1.
 6 
 7 Example:
 8 Input:
 9 [1,2,3]
10 Output:
11 3
12 Explanation:
13 Only three moves are needed (remember each move increments two elements):
14 [1,2,3]  =>  [2,3,3]  =>  [3,4,3]  =>  [4,4,4]
15 */
16 public class minMoves453 {
17     public static int minMoves(int[] nums) {
18         int min=nums[0];
19         int sum=0;
20         for(int i=1;i<nums.length;i++){
21             if(nums[i]<min)
22                 min=nums[i];
23         }
24         for(int i=0;i<nums.length;i++){
25             sum+=nums[i]-min;
26         }
27         return sum;
28     }
29     public static void main(String[] args) {
30         // TODO Auto-generated method stub
31         int[] nums={1,2,3};
32         int[] nums2={1,2,4,6};
33         System.out.println(minMoves(nums));
34         System.out.println(minMoves(nums2));
35     }
36 
37 }