1013 Partition Array Into Three Parts With Equal Sum 将数组分成和相等的三个部分
Description:
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:
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] <= 10000
题目描述:
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 :
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
思路:
模拟
找到是否有 3 段相等
时间复杂度O(n), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
bool canThreePartsEqualSum(vector& A)
{
int s = accumulate(A.begin(), A.end(), 0), n = A.size(), cur_sum = 0, count = 0;
if (s % 3) return false;
s /= 3;
for (int i = 0; i < n - 1; i++)
{
cur_sum += A[i];
if (cur_sum == s) {
++count;
if (count == 2) return true;
cur_sum = 0;
}
}
return false;
}
};
Java:
class Solution {
public boolean canThreePartsEqualSum(int[] A) {
int s = 0, n = A.length, curSum = 0, count = 0;
for (int a : A) s += a;
if (s % 3 != 0) return false;
s /= 3;
for (int i = 0; i < n - 1; i++) {
curSum += A[i];
if (curSum == s) {
++count;
if (count == 2) return true;
curSum = 0;
}
}
return false;
}
}
Python:
class Solution:
def canThreePartsEqualSum(self, arr: List[int]) -> bool:
if (s := sum(arr)) % 3:
return False
n, cur_sum, count = len(arr), 0, 0
s //= 3
for i in range(n - 1):
cur_sum += arr[i];
if cur_sum == s:
count += 1
if count == 2:
return True
cur_sum = 0
return False