Given a non-empty integer array of size n, find the minimum number of moves required to make all array elements equal, where a move is incrementing n - 1 elements by 1.
Example:
Input: [1,2,3] Output: 3 Explanation: Only three moves are needed (remember each move increments two elements): [1,2,3] => [2,3,3] => [3,4,3] => [4,4,4]
public class Solution { public int minMoves(int[] nums) { if (nums.length == 0) return 0; int min = nums[0]; for (int n : nums) min = Math.min(min, n); int res = 0; for (int n : nums) res += n - min; return res; } }
我日,这些人脑子里都在想什么?这答案也太brainteaser了
意思是:要给n-1位进行++,那我说我偏不,我就给所有n个数--,然后再给n-1个数++也可以吧?因为我们知道整体挪动不影响大局
那么问题来了,这是什么呢?这就是给1个数进行--,
然后问题就变成:只能一次对一个数进行--运算,要多少次才能让所有数相等。
最后就变成:找到minimum,让所有其他数向他靠拢
卧槽!