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 <= 5000 <= nums[i] <= 100
题目大意：给你一个数组nums，对于其中每个元素nums[i]，请你统计数组中比它小的所有数字的数目.也就是说，对于每个nums[i]你必须计算出有效的j的数量，其中j满足j != i 且 nums[j] < nums[i]. 以数组形式返回答案.
思路一：暴力法，对于每个nums[i], 统计数组中所有小于nums[i]的个数，时间复杂度$O(n^2)$.
C++代码：
1 class Solution { 2 public: 3 vector<int> smallerNumbersThanCurrent(vector<int>& nums) { 4 int len = nums.size(); 5 vector<int> cnt(len, 0); 6 for (int index = 0; index < len; ++index) { 7 for (int i = 0; i < len; ++i) { 8 if (i != index && nums[i] < nums[index]) 9 cnt[index]++; 10 } 11 } 12 return cnt; 13 } 14 };
python3代码：
思路二：由于数组中的数属于[0,100], 可以利用计数排序的方式，先将数组排好序。
C++代码：时间复杂度$O(n)$
class Solution { public: vector<int> smallerNumbersThanCurrent(vector<int>& nums) { vector<int> cnt(101, 0); for (int i = 0; i < nums.size(); ++i) {//计数， cnt[i]此时统计的是数组中数i的个数 cnt[nums[i]]++; } for (int i = 1; i < 101; ++i) {//cnt[i]统计的是数组中小于等于数i的个数 cnt[i] += cnt[i - 1]; } vector<int> ans(nums.size(), 0); for (int i = 0; i < nums.size(); ++i) { //如果nums[i] == 0, 说明数组中小于0的个数为0，否则小于nums[i]的个数为cnt[nums[i] - 1]; ans[i] = (nums[i] == 0 ? 0 : cnt[nums[i] - 1]); } return ans; } };
python3代码：
时间复杂度O(nlogn)
class Solution: def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]: indics = {} for index, num in enumerate(sorted(nums)): indics.setdefault(num, index) return [indics[num] for num in nums]
时间复杂度O(n):
class Solution: def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]: count = collections.Counter(nums) for i in range(1,101): count[i] += count[i-1] return [count[x-1] for x in nums]