LeetCode-453. Minimum Moves to Equal Array Elements
2016-12-14 09:53
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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.
大意就是:给一个长度为n的非空整形数组,请找出使数组中所有元素相等的最小操作次数,每次操作可以给n-1个数加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]
其实给n-1个数字加1,等同于给那个未被选中的数字减1,比如数组[1,2,3], 给除去最大值的其他数字加1,变为[2,3,3],我们全体减1,并不影响数字间相对差异,变为[1,2,2],这个结果其实就是原始数组的最大值3自减1,那么问题也可能转化为,将所有数字都减小到最小值,然后累加每个数跟最小值之间的差值即可。
所以先排序,第一个数肯定是最小的,然后依次求差累加。
大意就是:给一个长度为n的非空整形数组,请找出使数组中所有元素相等的最小操作次数,每次操作可以给n-1个数加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]
class Solution { public: int minMoves(vector<int>& nums) { sort(nums.begin(),nums.end()); vector<int>::size_type sz = nums.size(); int move = 0; int target = nums[0]; for (int i=1;i<sz;i++) move = move + (nums[i]-target); return move; } };
其实给n-1个数字加1,等同于给那个未被选中的数字减1,比如数组[1,2,3], 给除去最大值的其他数字加1,变为[2,3,3],我们全体减1,并不影响数字间相对差异,变为[1,2,2],这个结果其实就是原始数组的最大值3自减1,那么问题也可能转化为,将所有数字都减小到最小值,然后累加每个数跟最小值之间的差值即可。
所以先排序,第一个数肯定是最小的,然后依次求差累加。
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