112th LeetCode Weekly Contest 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

问的是把数组的数字+1，几次后，数组的数字唯一了。

这里有个解法，把重复的拿出来，从小到大排序。从[min,max]（max不一定就是代码的那个），哪个位置缺了就填哪个，然后算sum +=(i-ans);

其实优雅的解法应该是第二个

class Solution {

public:

    int minIncrementForUnique(vector<int>& A) {

        int sum = ;

        int ans = ;

        int pos = ;

        int Min = ;

        map<int,int> mp;

        stack<int>st;

        vector<int>ve;

        for(int i=;i<A.size();i++){

            Min = min(Min,A[i]);

            mp[A[i]]++;

            if(mp[A[i]]>=){

                ve.push_back(A[i]);

            }

        }

        sort(ve.begin(),ve.end());

        for(int i=ve.size()-;i>=;i--){

            st.push(ve[i]);

        }

        for(int i=;i<;i++){

            if(st.empty()){

                break;

            }

            if(mp[i]==){

                ans = st.top();

                if(i<ans){

                    continue;

                }

                sum +=(i-ans);

                st.pop();

                mp[i]=;

            }

        }

        return sum;

    }

};

class Solution {

public:

    int minIncrementForUnique(vector<int>& A) {

        sort(A.begin(), A.end());

        int lowest = -, total = ;

        for (int a : A) {

            lowest = max(lowest, a);

            total += lowest - a;

            lowest++;

        }

        return total;

    }

};

巴特西

112th LeetCode Weekly Contest Minimum Increment to Make Array Unique

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