1013. Partition Array Into Three Parts With Equal Sum
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: true
Explanation: 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: true
Explanation: 3 + 3 = 6 = 5 - 2 + 2 + 5 + 1 - 9 + 4
Constraints:
3 <= A.length <= 50000-10^4 <= A[i] <= 10^4
题目链接：https://leetcode-cn.com/problems/partition-array-into-three-parts-with-equal-sum/submissions/
思路
题目需要的是连续的列，所以双指针即可。
注意边界条件的设定。
以及有几个特殊样例：
[1,-1,-1,1]
[10,-10,10,-10,10,-10,10,-10,]class Solution {
public:
bool canThreePartsEqualSum(vector<int>& A) {
int num = A.size();
if(num<=2) return false;
int sum = accumulate(A.begin(),A.end(), 0);
if(sum%3!=0) return false;
int l = 0, r = num-1;
int lsum = A[l++], rsum = A[r--];
sum /= 3;
while(l<r){
if(lsum==sum && rsum==sum){
break;
}
if(lsum!=sum){
lsum += A[l++];
}
if(rsum!=sum){
rsum += A[r--];
}
}
if(lsum==sum && rsum==sum && l<r+1 ) return true;
else return false;
}
};