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: 1
Explanation: After 1 move, the array could be [1, 2, 3].
Example 2:
Input: [3,2,1,2,1,7]
Output: 6
Explanation: 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:
0 <= A.length <= 40000
0 <= A[i] < 40000
给定整数数组 A,每次 move 操作将会选择任意 A[i]
,并将其递增 1
。
返回使 A
中的每个值都是唯一的最少操作次数。
示例 1:
输入:[1,2,2] 输出:1 解释:经过一次 move 操作,数组将变为 [1, 2, 3]。
示例 2:
输入:[3,2,1,2,1,7] 输出:6 解释:经过 6 次 move 操作,数组将变为 [3, 4, 1, 2, 5, 7]。 可以看出 5 次或 5 次以下的 move 操作是不能让数组的每个值唯一的。
提示:
0 <= A.length <= 40000
0 <= A[i] < 40000
508ms
1 class Solution { 2 func minIncrementForUnique(_ A: [Int]) -> Int { 3 var arr:[Int] = A.sorted(by:<) 4 var r:Int = -1 5 var ret:Int = 0 6 for i in 0..<arr.count 7 { 8 var to:Int = max(r,arr[i]) 9 ret += abs(to - arr[i]) 10 r = to + 1 11 } 12 return ret 13 } 14 }