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🐱(array): 1013. 将数组分成和相等的三个部分

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docs/data-structure/array
- README.md

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`@@ -2663,4 +2663,49 @@ class Solution(object):`
`2663`	`2663`	`return tmp_string`
`2664`	`2664`	```
`2665`	`2665`
	`2666`	`+## 1013. 将数组分成和相等的三个部分`
`2666`	`2667`
	`2668`	`+[原题链接](https://leetcode-cn.com/problems/partition-array-into-three-parts-with-equal-sum/)`
	`2669`	`+`
	`2670`	`+### 思路`
	`2671`	`+`
	`2672`	+1. 数组求和,看是否可以整除三,不能整除则直接返回 `false`
	`2673`	`+2. 双指针指向数组前后,前后都分别开始找累加和等于目标值的部分,如果前后都能找到,则中间的自动就找到了`
	`2674`	`+`
	`2675`	`+注意问题:`
	`2676`	`+`
	`2677`	+1. 存在 `[1,-1,1,-1]` 这样的用例,需要判断双指针最终位置,相距是否大于 1,即中间是否还有位置
	`2678`	+2. 目标值可以为 0,所以左右区间的数据应当初始化为 `A[0]` 与 `A[length - 1]`
	`2679`	`+`
	`2680`	+```python
	`2681`	`+class Solution:`
	`2682`	`+ def canThreePartsEqualSum(self, A: List[int]) -> bool:`
	`2683`	`+ # 求总和`
	`2684`	`+ s = sum(A)`
	`2685`	`+ if s % 3 != 0:`
	`2686`	`+ return False`
	`2687`	`+ part = s // 3`
	`2688`	`+`
	`2689`	`+ # 双指针`
	`2690`	`+ i = 0`
	`2691`	`+ j = len(A) - 1`
	`2692`	`+ # 计算左右区间和`
	`2693`	`+ left_part = A[i]`
	`2694`	`+ right_part = A[j]`
	`2695`	`+ res = False`
	`2696`	`+ while i < j:`
	`2697`	`+ if left_part != part:`
	`2698`	`+ i += 1`
	`2699`	`+ left_part += A[i]`
	`2700`	`+`
	`2701`	`+ if right_part != part:`
	`2702`	`+ j -= 1`
	`2703`	`+ right_part += A[j]`
	`2704`	`+`
	`2705`	`+ if left_part == part and right_part == part:`
	`2706`	`+ res = True`
	`2707`	`+ break`
	`2708`	`+ # print(i, j)`
	`2709`	`+ # 处理特殊情况:[1,-1,1,-1]`
	`2710`	`+ return res and j - i > 1`
	`2711`	+```

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