equilibrium index问题复杂度O(N)的解决方法

public class Solution {
//平衡索引算法-java
public int solution(int[] arr){

int sumright=0;
int sumleft=0;
int index=0;
for(int i=0;i<arr.length;i++){
sumright+=arr[i];
}

for(int i=0;i<arr.length;i++){
sumleft+=arr[i];
sumright-=arr[i];
if(sumleft==sumright){
return i;
}else if(sumleft>sumright){
return -1;
}
}
return -1;
}
}
//在主方法中调用：
int[] arr={5,7,1,3,4,3,7,2};
System.out.print(new Solution().solution(arr));

class Solution:
#平衡索引算法-python
def slotion(self,arr):
sumleft = 0
sumright = 0
for item in arr:
sumright = sumright + item

for item in arr:
sumleft = sumleft + item
sumright = sumright - item
if sumleft == sumright:
return arr.index(item)
elif sumleft > sumright:
return -1

#arr=[5,7,1,1,4,3,5,2]
#print(Solution().slotion(arr))

namespace algorithm
{
public class Solution
{
//平衡索引算法-c#
public int solution(int[] arr)
{
int sumright = 0;
int sumleft = 0;
for (int i = 0; i < arr.Count(); i++)
{
sumright += arr[i];
}

for (int i = 0; i < arr.Count(); i++)
{
sumleft += arr[i];
sumright -= arr[i];
if (sumleft == sumright)
{
return i;
}
else if (sumleft > sumright)
{
return -1;
}
}
return -1;
}
}
}
//在主方法中调用：
int[] arr={5,7,1,3,4,3,7,2};
Console.WriteLine(new Solution().solution(arr));
Console.Read();