leetcode-1365-easy

How Many Numbers Are Smaller Than the Current Number

Given the array nums, for each nums[i] find out how many numbers in the array are smaller than it. That is, for each nums[i] you have to count the number of valid j's such that j != i and nums[j] < nums[i].

Return the answer in an array.

Example 1:

Input: nums = [8,1,2,2,3]
Output: [4,0,1,1,3]
Explanation: 
For nums[0]=8 there exist four smaller numbers than it (1, 2, 2 and 3). 
For nums[1]=1 does not exist any smaller number than it.
For nums[2]=2 there exist one smaller number than it (1). 
For nums[3]=2 there exist one smaller number than it (1). 
For nums[4]=3 there exist three smaller numbers than it (1, 2 and 2).
Example 2:

Input: nums = [6,5,4,8]
Output: [2,1,0,3]
Example 3:

Input: nums = [7,7,7,7]
Output: [0,0,0,0]
Constraints:

2 <= nums.length <= 500
0 <= nums[i] <= 100

思路一：先排序，然后从小到大统计数量

    public static int[] smallerNumbersThanCurrent(int[] nums) {
        int[] result = new int[nums.length];
        int[] clone = nums.clone();
        Arrays.sort(nums);
        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 1; i < nums.length; i++) {
            if (nums[i] > nums[i - 1]) {
                map.put(nums[i], i);
            }
        }

        for (int i = 0; i < result.length; i++) {
            result[i] = map.getOrDefault(clone[i], 0);
        }

        return result;
    }