o(n2)和o(n)
//时间复杂度是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]); } } } //时间复杂度是n2 static void equPoint1(){ int[] arr = {2,3,4,5,8,5,9}; for (int i = 0; i < arr.length; i++) { //计算每个数字的前后和 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; }