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] <= 10000class 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; } }
(Easy) Partition Array Into Three Parts With Equal Sum - LeetCode
原文：https://www.cnblogs.com/codingyangmao/p/11468713.html