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多版本代码实现
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ztianming authored Jul 29, 2020
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![](../Animation/Animation.gif)

### 代码实现

#### C++
```
// 1. Two Sum
// https://leetcode.com/problems/two-sum/description/
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};

```




#### C
```c
// 1. Two Sum
// https://leetcode.com/problems/two-sum/description/
// 时间复杂度:O(n)
// 空间复杂度:O(n)
/**
* Note: The returned array must be malloced, assume caller calls free().
*/
int* twoSum(int* nums, int numsSize, int target, int* returnSize){
int *ans=(int *)malloc(2 * sizeof(int));
int i,j;
bool flag=false;
for(i=0;i<numsSize-1;i++)
{
for(j=i+1;j<numsSize;j++)
{
if(nums[i]+nums[j] == target)
{
ans[0]=i;
ans[1]=j;
flag=true;
}
}
}
if(flag){
*returnSize = 2;
}
else{
*returnSize = 0;
}
return ans;
}
```
#### Java
```
// 1. Two Sum
// https://leetcode.com/problems/two-sum/description/
// 时间复杂度:O(n)
// 空间复杂度:O(n)
class Solution {
public int[] twoSum(int[] nums, int target) {
int l = nums.length;
int[] ans=new int[2];
int i,j;
for(i=0;i<l-1;i++)
{
for(j=i+1;j<l;j++)
{
if(nums[i]+nums[j] == target)
{
ans[0]=i;
ans[1]=j;
}
}
}

return ans;

}
}
```
#### Python
```
# 1. Two Sum
# https://leetcode.com/problems/two-sum/description/
# 时间复杂度O(n)
# 空间复杂度O(n)
class Solution(object):
def twoSum(self, nums, target):
l = len(nums)
print(nums)
ans=[]
for i in range(l-1):
for j in range(i+1,l):
if nums[i]+nums[j] == target:
ans.append(i)
ans.append(j)
print([i,j])
break
return ans
```

![](../../Pictures/qrcode.jpg)

5 comments on commit d2b72e5

@zhenpeiwang
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Hi, Thank you for sharing the codes. The time complexity of the codes here seems to be O(n^2) instead of O(n). Please correct me if i am wrong.

@Shirman
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@Shirman Shirman commented on d2b72e5 Jan 1, 2022 via email

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@archerasdo
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@archerasdo archerasdo commented on d2b72e5 Jan 1, 2022 via email

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@Teemo100
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@Teemo100 Teemo100 commented on d2b72e5 Jan 1, 2022 via email

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@HongXiaoHong
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@HongXiaoHong HongXiaoHong commented on d2b72e5 Jan 1, 2022 via email

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