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1013.将数组分成和相等的三个部分.c
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1013.将数组分成和相等的三个部分.c
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/*
* @lc app=leetcode.cn id=1013 lang=c
*
* [1013] 将数组分成和相等的三个部分
*
* https://leetcode-cn.com/problems/partition-array-into-three-parts-with-equal-sum/description/
*
* algorithms
* Easy (40.37%)
* Likes: 118
* Dislikes: 0
* Total Accepted: 36.8K
* Total Submissions: 91.2K
* Testcase Example: '[0,2,1,-6,6,-7,9,1,2,0,1]'
*
* 给你一个整数数组 A,只有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:
*
* 输入:[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
* -10^4 <= A[i] <= 10^4
*
*
*/
// @lc code=start
bool canThreePartsEqualSum(int* A, int ASize){
}
// @lc code=end