# leetcode 1365. How Many Numbers Are Smaller Than the Current Number

2019/04/10 10:10

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

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：

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：

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]

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]

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