> For the complete documentation index, see [llms.txt](https://dfine.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://dfine.gitbook.io/leetcode/1013.partition_array_into_three_parts_with_equal_sum.md).

# 1013.Partition Array Into Three Parts with Equal Sum

**1013.Partition Array Into Three Parts with Equal Sum**

难度:Easy

> 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。

```
示例 1：

输出：[0,2,1,-6,6,-7,9,1,2,0,1]
输出：true
解释：0 + 2 + 1 = -6 + 6 - 7 + 9 + 1 = 2 + 0 + 1
示例 2：

输入：[0,2,1,-6,6,7,9,-1,2,0,1]
输出：false
示例 3：

输入：[3,3,6,5,-2,2,5,1,-9,4]
输出：true
解释：3 + 3 = 6 = 5 - 2 + 2 + 5 + 1 - 9 + 4
 

提示：

3 <= A.length <= 50000
-10000 <= A[i] <= 10000
```

直接遍历，碰到前面两个和为sum/3的序列判断结束后，看最后剩下的和是不是sum/3。

```
class Solution {
public:
    bool canThreePartsEqualSum(vector<int>& A) {
        int sum=accumulate(A.begin(),A.end(),0);
       // cout<<sum<<endl;
        if(sum%3)
        return false;
           int start=0;
        for(int i=0;i<3;i++){
     
        int s=0;
        for( ;start<A.size();start++)
        {
            s+=A[start];
            if(s==sum/3) { ++start; break;}
        }
          //  cout<<s<< "   "<< start<<endl;
            if(s!=sum/3) return false;
        }
        return true;
    }
};
```

> 执行用时 :96 ms, 在所有 C++ 提交中击败了35.54%的用户\
> 内存消耗 :12.6 MB, 在所有 C++ 提交中击败了22.02%的用户


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