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:

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

解題思路

要能夠分為三部分，首先需要總和能夠被3整除。然后如果保證左側(cè)的總和跟中間的總和都是等于數(shù)組總和的三分之一，就說明可以分為三部分。

實現(xiàn)代碼

// Runtime: 1 ms, faster than 100.00% of Java online submissions for Partition Array Into Three Parts With Equal Sum.
// Memory Usage: 50.7 MB, less than 100.00% of Java online submissions for Partition Array Into Three Parts With Equal Sum.
class Solution {
    public boolean canThreePartsEqualSum(int[] A) {
        int sum = 0;
        for (int i = 0; i < A.length; i++) {
            sum += A[i];
        }
        
        if (sum % 3 != 0) {
            return false;
        }
        
        int leftSum = 0, midSum = 0;
        for (int i = 0; i < A.length; i++) {
            if (leftSum == sum / 3) {
                if (midSum == sum / 3) {
                    return true;
                } else {
                    midSum += A[i];
                }
            } else {
                leftSum += A[i];
            }
        }
        
        return false;
    }
}