945.Minimum Increment to Make Array Unique

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

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/minimum-increment-to-make-array-unique
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框架

class Solution {
public:
    int minIncrementForUnique(vector<int>& A) {

    }
};

1. 计数`O(n+range)`

数据是[0, 40000)，所以移动之后占的位置是[0, 79998].
但是我们并不需要80000大的数组，40001就足够了，因为39999位置上的次数-1加到40000位置上，这些数就被均匀地分布到40000~之后了。
同理，我们可以记录maxv，到达maxv之后，后面的maxv+1, maxv+2, ...用公式算就好了。

class Solution {
public:
    int minIncrementForUnique(vector<int>& A) {
        int count[40010];
        for (int i = 0; i < 40010; i++)
            count[i] = 0;
        int maxv = 0;
        int ans = 0;
        for (int i = 0; i < A.size(); i++) {
            count[A[i]]++;
            maxv = maxv >= A[i] ? maxv : A[i];
        }
        
        for (int i = 0; i <= maxv; i++) {
            if (count[i] > 1) {
                count[i + 1] += count[i] - 1;
                ans += count[i] - 1;
                count[i] = 1;
            }
        }

        ans += count[maxv + 1] * (count[maxv + 1] - 1) / 2;

        return ans;
    }
};

框架

1. 计数O(n+range)

1. 计数`O(n+range)`