【數組】C063_LC_有多少小於當前數字的數字（思維題 = 排序 + mp / 桶排序）

一、Problem

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.

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).

二、Solution

方法一：排序

• Brute force 不難想，但也卻是存在更優解。
• 比如說當 nums 是有序的情況下，那麼最後一個元素 b 肯定是最大的，那麼 b 的前面的所有元素都比 b 小。

class Solution {
    public int[] smallerNumbersThanCurrent(int[] nums) {
        int N = nums.length, a[] = new int[N];
        a = Arrays.copyOf(nums, N);
        Arrays.sort(a);
        Map<Integer, Integer> mp = new HashMap<>();
        for (int i = 0; i < N; i++)
            mp.putIfAbsent(a[i], i);
        for (int i = 0; i < N; i++)
            nums[i] = mp.get(nums[i]);
        return nums;
    }
}

複雜度分析

• 時間複雜度：$O(nlogn)$，
• 空間複雜度：$O(n)$，

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方法二：桶排序

• 最簡單的桶排序就是用一個整形數組 int[] freq 對元素進行計數，然後如果某個元素 a 出現頻次爲 n 次，則將 a 輸出 n 次
• 在這提應用起來就是：某個桶 freq[i] 前面所有元素的頻次就是有多少元素小於當前元素。

class Solution {
    public int[] smallerNumbersThanCurrent(int[] nums) {
        int N = nums.length, freq[] = new int[100+5];
        for (int n : nums) freq[n]++;
        for (int i = 1; i < freq.length; i++) freq[i] += freq[i-1];

        int[] res = new int[N];
        for (int i = 0; i < N; i++) {
            if (nums[i] > 0)
                res[i] = freq[nums[i]-1];
        }
        return res;
    }
}

複雜度分析

• 時間複雜度：$O(n)$，
• 空間複雜度：$O(...)$，