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:

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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]

为了让n个数相等, n - 1个数加上1, 相当于最大的一个数减去1.
所以最好的让所有数相等于方式是让所有的数经过x次减法后都等于min, 需要的步骤数为sum - n * min.

代码如下:

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public int minMoves(int[] nums) {
    if (nums == null || nums.length == 0)
        return 0;
    int min = Integer.MAX_VALUE;
    int sum = 0;
    for (int n : nums) {
        sum += n;
        min = Math.min(min, n);
    }
    return sum - nums.length * min;
}