[leetcode-453-Minimum Moves to Equal Array Elements]
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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.
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]
思路:
参考自:https://discuss.leetcode.com/topic/66737/it-is-a-math-question
let‘s define sum as the sum of all the numbers, before any moves; minNum as the min number int the list; n is the length of the list;
After, say m moves, we get all the numbers as x , and we will get the following equation
sum + m * (n - 1) = x * n
and actually,
x = minNum + m
and finally, we will get
sum - minNum * n = m
So, it is clear and easy now.
int minMoves(vector<int>& nums) { int ret = 0; int smallest = *min_element(nums.begin(), nums.end()); for (auto a:nums) { ret += (a - smallest); } return ret; }
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