(Easy) Partition Array Into Three Parts With Equal Sum - LeetCode

Description:

Given an array A of integers, return true if and only if we can partition the array into three non-empty parts with equal sums.

Formally, we can partition the array if we can find indexes i+1 < j with (A[0] + A[1] + ... + A[i] == A[i+1] + A[i+2] + ... + A[j-1] == A[j] + A[j-1] + ... + A[A.length - 1])

Example 1:

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

Example 2:

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

Example 3:

Input: [3,3,6,5,-2,2,5,1,-9,4]
Output: true
Explanation: 3 + 3 = 6 = 5 - 2 + 2 + 5 + 1 - 9 + 4

Note:

1. 3 <= A.length <= 50000
2. -10000 <= A[i] <= 10000

Accepted

18,270

Submissions

32,514

Solution:

class Solution {
    public boolean canThreePartsEqualSum(int[] A) {

        // int Sum of each part
        int sum = 0;

        for(int i = 0; i< A.length; i++){

            sum = sum + A[i];
        }

        sum = sum / 3; 

        int find_i =-1;
        int find_j = -1;
        int cur_sum = 0;
        for(int i = 0; i< A.length; i++){

            cur_sum = cur_sum +A[i];

            if(cur_sum == sum){

                find_i = i;

                cur_sum = 0;
                break;
            }
        }

        for(int k = find_i+1; k<A.length; k++){
            cur_sum = cur_sum+A[k];

            if(cur_sum ==sum){

                find_j = k;
                break;
            }

        }

       if(find_i <find_j && find_i>=0&& find_j <A.length){
           return true;
       }
      return false;
    }
}

原文地址：https://www.cnblogs.com/codingyangmao/p/11468713.html