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Merge pull request #2850 from yyccPhil/master
feat: Updated题目1365,提供了Python的暴力解法,以及使用数组进行哈希的解法
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@ -115,7 +115,7 @@ public:
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## 其他语言版本
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### Java:
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### Java:
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```Java
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public int[] smallerNumbersThanCurrent(int[] nums) {
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@ -138,18 +138,51 @@ public int[] smallerNumbersThanCurrent(int[] nums) {
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### Python:
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```python
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> 暴力法:
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```python3
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class Solution:
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def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
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res = [0 for _ in range(len(nums))]
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for i in range(len(nums)):
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cnt = 0
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for j in range(len(nums)):
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if j == i:
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continue
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if nums[i] > nums[j]:
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cnt += 1
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res[i] = cnt
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return res
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```
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> 排序+hash:
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```python3
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class Solution:
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# 方法一:使用字典
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def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
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res = nums[:]
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hash = dict()
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hash_dict = dict()
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res.sort() # 从小到大排序之后,元素下标就是小于当前数字的数字
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for i, num in enumerate(res):
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if num not in hash.keys(): # 遇到了相同的数字,那么不需要更新该 number 的情况
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hash[num] = i
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if num not in hash_dict.keys(): # 遇到了相同的数字,那么不需要更新该 number 的情况
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hash_dict[num] = i
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for i, num in enumerate(nums):
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res[i] = hash[num]
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res[i] = hash_dict[num]
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return res
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# 方法二:使用数组
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def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
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# 同步进行排序和创建新数组的操作,这样可以减少一次冗余的数组复制操作,以减少一次O(n) 的复制时间开销
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sort_nums = sorted(nums)
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# 题意中 0 <= nums[i] <= 100,故range的参数设为101
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hash_lst = [0 for _ in range(101)]
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# 从后向前遍历,这样hash里存放的就是相同元素最左面的数值和下标了
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for i in range(len(sort_nums)-1,-1,-1):
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hash_lst[sort_nums[i]] = i
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for i in range(len(nums)):
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nums[i] = hash_lst[nums[i]]
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return nums
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```
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### Go:
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@ -220,7 +253,7 @@ var smallerNumbersThanCurrent = function(nums) {
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};
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```
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### TypeScript:
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### TypeScript:
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> 暴力法:
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@ -241,7 +274,7 @@ function smallerNumbersThanCurrent(nums: number[]): number[] {
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};
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```
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> 排序+hash
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> 排序+hash:
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```typescript
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function smallerNumbersThanCurrent(nums: number[]): number[] {
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@ -260,7 +293,7 @@ function smallerNumbersThanCurrent(nums: number[]): number[] {
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};
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```
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### rust
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### Rust:
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```rust
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use std::collections::HashMap;
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impl Solution {
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