原始题目

Given an array of integers arr, return true 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
(arr[0] + arr[1] + ... + arr[i] == arr[i + 1] + arr[i + 2] + ... + arr[j - 1] == arr[j] + arr[j + 1] + ... + arr[arr.length - 1])

Example 1:

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

Example 3:

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

Constraints:

• 3 <= arr.length <= 5 * 104
• -104 <= arr[i] <= 104

题目分析

一个阵列，当里面的内容相加可以分成完全相等的三个部分，并且内容都不是空值就会传 true
否则回传 false

解题过程

可以理解成三等分的每一份恰好等於阵列总和的三分之一，所以不能整除三的就必定是 false
从阵列开头累加，当等於三分之一的总和，则切为一份
继续计算下一份的累加值，当累加值等於三分之二的总和时，剩下的值总和必然是另外的三分之一，所以可以回传 true
若累加没有达到三分之二则回传 false

class Solution:
    def canThreePartsEqualSum(self, arr: List[int]) -> bool:
        s = sum(arr)
        if s % 3 != 0:
            return False
        
        target = s // 3
        n, i, cur = len(arr), 0, 0
        
        while i < n:
            cur += arr[i]
            if cur == target:
                break
            i += 1
        
        if cur != target:
            return False
        
        j = i + 1
        while j + 1 < n:
            cur += arr[j]
            if cur == target * 2:
                return True
            j += 1
            
        return False

结果