[leetcode] 1013. Partition Array Into Three Parts With Equal Sum

[leetcode] 1013. Partition Array Into Three Parts With Equal Sum

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: A = [0,2,1,-6,6,-7,9,1,2,0,1]Output: trueExplanation: 0 + 2 + 1 = -6 + 6 - 7 + 9 + 1 = 2 + 0 + 1

Example 2:

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

Example 3:

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

Note:

3 <= A.length <= 50000.-10^4 <= A[i] <= 10^4.

分析

题目的意思是：给你一个数组，判断该数组能否被分为三个相等的子数组。我的思路也很简单，就先求出相等的值是多少，然后再去找该数组可以被划分为多少个这样的相等值的数组，如果小于3个，则返回True，如果大于3个，则要判断能否被3整除，如果不能则判断是否为0.

代码

class Solution: def canThreePartsEqualSum(self, A: List[int]) -> bool: sum_A=sum(A) val=sum_A//3 if(val !=sum_A/3): return False n=len(A) t=0 list_cnt=[] for i in range(n): t+=A[i] if(t==val): list_cnt.append(t) t=0 else: continue if(len(list_cnt)>=3): if(list_cnt[0]==0): return True elif(len(list_cnt)==3): return True return False

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