LeetCode 32. Make the array only the smallest increment

Title Description

Given an array of integers A, select any operation will move each A [i], and increments its 1.

A return value of each of the minimum number of operations is unique.

Example 1:

Input: [1,2,2]
Output: 1
Explanation: Following a move operation, the array becomes [1, 2, 3].
Example 2:

Input: [3,2,1,2,1,7]
Output: 6
Explanation: after 6 move operation, the array becomes [3, 4, 1, 2, 5, 7].
The following can be seen five times or five times move operation is not possible for each unique value of the array.
prompt:

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

 

Problem-solving ideas

1.先排序

2.遍历数组，若当前元素小于等于它的前一个元素，则将其变为前一个数+1

3.复杂度O(nlogn)

code show as below

package leetcode;

import java.util.Arrays;

public class MinIncrementForUnique {
    
     
     public int minIncrementForUnique(int[] A) {
         int result=0; 
         Arrays.sort(A);
         for (int i = 0; i < A.length; i++) {
            if (A[i]<=A[i-1]) {
                int pre=A[i];
                A[i]=A[i-1]+1;
                result+=A[i]-pre;
            }
        }
         return result;

        }
     
}