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])
给定一个整数数组 A，只有我们可以将其划分为三个和相等的非空部分时才返回 true，否则返回 false。
形式上，如果我们可以找出索引 i+1 < j 且满足 (A[0] + A[1] + ... + A[i] == A[i+1] + A[i+2] + ... + A[j-1] == A[j] + A[j-1] + ... + A[A.length - 1]) 就可以将数组三等分。
题目链接：https://leetcode.com/problems/partition-array-into-three-parts-with-equal-sum/
Difficulty: easy
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
分析
- 若和不能被3整除，返回False，tar=sum(A)/3;
- index记录当前遍历所有的数字的和，s为和为tar的序列数；
- 遍历A，index加上当前数字，若index等于tar，s加一，index置零；
- 若s等于3返回True，反之False。
参考代码
class Solution(object):
def canThreePartsEqualSum(self, A):
N=len(A)
tar=sum(A)//3
if(tar*3!=sum(A)):
return False
index=0
s=0
for i in range(N):
index+=A[i]
if(index==tar):
index=0
s+=1
if(s==3):
return True
return False