The array is divided into three equal portions

problem

To give you an array of integers A, can only return it into three equal and non-null only part true, otherwise false.
Formally, if we can find the index i + 1 <j satisfying (A [0] + A [ 1] + ... + A [i] == A [i + 1] + A [i + 2] + ... + A [j-1] == A [j] + A [j-1] + ... + A [A.length - 1]) can be an array of three aliquots.

Examples

Example 1:

Output: [0,2,1, -6,6, -7,9,1,2,0,1]
Output: true
interpretation: 2 + 0 + 1 = -6 + 6--7 = 2 + 9 + 1 + 0 + 1

Example 2:

Input: [0,2,1, -6,6,7,9, -1,2,0,1]
Output: false

Problem-solving ideas

1. Since the three parts is not empty, so when A.length <3, returns to false;
2. summing array, if the modulo 3 and p is not zero, return to false;
3. Make the array three equal parts, and in addition to the array 3, and each part is obtained, it is determined to continue, and if each part has a value equal to the value, returns true, otherwise returns false;

Problem-solving Code

class Solution {
    public boolean canThreePartsEqualSum(int[] A) {
        if(A.length<3)
            return false;
        int sum=0;
        for(int i=0;i<A.length;i++){
            sum=sum+A[i];
        }
        int a=0;
        if(sum!=0){
            if(sum%3!=0)
                return false;
            a=sum/3;
        }
         int b=0;
        int c=0;
        for(int j=0;j<A.length;j++){
            b=b+A[j];
            if(b==a && c<2){
                c++;
                b=0;
                if(j==(A.length-1))
                    return false;
            }
        }
        if(c==2 && b==a)
            return true;
        else
            return false;


    }
}
``