1365. How Many Numbers Are Smaller Than the Current Number (E)

How Many Numbers Are Smaller Than the Current Number (E)

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

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题意

对于一个数组nums中的每一个数nums[i]，统计小于nums[i]的数的个数。

思路

1. 暴力法。
2. Hash。
3. 二分搜索。

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代码实现 - 暴力法

class Solution {
    public int[] smallerNumbersThanCurrent(int[] nums) {
        int[] ans = new int[nums.length];
        for (int i = 0; i < nums.length; i++) {
            for (int j = 0; j < nums.length; j++) {
                if (j != i && nums[j] < nums[i]) {
                    ans[i]++;
                }
            }
        }
        return ans;
    }
}

代码实现 - Hash

class Solution {
    public int[] smallerNumbersThanCurrent(int[] nums) {
        int[] ans = new int[nums.length];
        Map<Integer, Integer> hash = new TreeMap<>();
        for (int i = 0; i < nums.length; i++) {
            if (!hash.containsKey(nums[i])) {
                hash.put(nums[i], 1);
            } else {
                hash.put(nums[i], hash.get(nums[i]) + 1);
            }
        }
        int sum = 0;
        for (int key : hash.keySet()) {
            int temp = sum + hash.get(key);
            hash.put(key, sum);
            sum = temp;
        }
        for (int i = 0; i < ans.length; i++) {
            ans[i] = hash.get(nums[i]);
        }
        return ans;
    }
}

代码实现 - 二分法

class Solution {
    public int[] smallerNumbersThanCurrent(int[] nums) {
        int[] ans = new int[nums.length];
        int[] copy = Arrays.copyOf(nums, nums.length);
        Arrays.sort(copy);
        for (int i = 0; i < ans.length; i++) {
            ans[i] = binarySearch(copy, nums[i]);
        }
        return ans;
    }

    private int binarySearch(int[] nums, int target) {
        int left = 0, right = nums.length - 1;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] < target) {
                left = mid + 1;
            } else if (nums[mid] >= target) {
                right = mid;
            }
        }
        return left;
    }
}