尋找平衡點的倆種方法

o(n2)和o(n)

//時間複雜度是n
static void equPoint2(){
        int[]  arr =   {2,3,4,5,8,5,9};
        int leftSum = arr[0];
        int rightSum = 0;
        for (int i : arr) {
            rightSum+=i;
        }

        for (int i =1; i < arr.length; i++) {
            leftSum+=arr[i-1];
            rightSum-=arr[i];
            if(leftSum==rightSum){
                System.out.println(arr[i]);
            }
        }
    }

//時間複雜度是n2
    static void equPoint1(){
        int[]  arr =   {2,3,4,5,8,5,9};
        for (int i = 0; i < arr.length; i++) {
            //計算每個數字的前後和
            if(isOk(arr, i)){
                System.out.println(arr[i]);
            }
        }
    }
    static boolean isOk(int[] arr,int  position){
        int leftSum =  0 ;
        int rightSum =  0 ;

        for (int i = 0; i <position; i++) {
            leftSum+=arr[i];
        }
        for(int i = position+1;i<arr.length;i++){
            rightSum +=arr[i];
        }

        return leftSum==rightSum;
    }