题目如下：

Given an array of integers A, a move consists of choosing any A[i], and incrementing it by 1.

Return the least number of moves to make every value in A unique.

 

Example 1:

Input: [1,2,2]Output: 1Explanation:  After 1 move, the array could be [1, 2, 3].

Example 2:

Input: [3,2,1,2,1,7]Output: 6Explanation:  After 6 moves, the array could be [3, 4, 1, 2, 5, 7].It can be shown with 5 or less moves that it is impossible for the array to have all unique values. 

Note:

1. 0 <= A.length <= 40000
2. 0 <= A[i] < 40000

解题思路：本题的解法类似贪心算法。首先找出所有重复的值并且按升序排好，对于每个重复的值，都找出最小的并且大于自身以及当前不与其他值重复的值，两者的差值即为move的步数。

代码如下：

class Solution(object):    def minIncrementForUnique(self, A):        """        :type A: List[int]        :rtype: int        """        dic = {}        dup = []        for i in A:            if i in dic:                dup.append(i)            else:                dic[i] = 1        dup.sort()        res = 0        current = 0        for i in dup:            current = max(i+1,current)            while True:                if current in dic:                    current += 1                    continue                dic[current] = 1                res += (current - i)                current += 1                break        return res