Last Updated on 2022-11-28 by Clay
題目
Given a positive integer n
, find the pivot integer x
such that:
- The sum of all elements between
1
andx
inclusively equals the sum of all elements betweenx
andn
inclusively.
Return the pivot integer x
. If no such integer exists, return -1
. It is guaranteed that there will be at most one pivot index for the given input.
Example 1:
Input: n = 8 Output: 6 Explanation: 6 is the pivot integer since: 1 + 2 + 3 + 4 + 5 + 6 = 6 + 7 + 8 = 21.
Example 2:
Input: n = 1 Output: 1 Explanation: 1 is the pivot integer since: 1 = 1.
Example 3:
Input: n = 4 Output: -1 Explanation: It can be proved that no such integer exist.
Constraints:
1 <= n <= 1000
題目給定一個從 1…n 的公差為 1 的等差數列,而我們要在數列中新增一個數值 x
,並且可以讓 1+...+x
剛好等於 x+...+n
。
所以 x
這個數值又被稱為 pivot integer。
解題思路
我不太確定有沒有更好的解法,不過基本上我就是把 x=n
一路計算到 x=1
(不過在確認 x+…+n 肯定比另外一邊大的時候我就會返回 -1、表示沒有解了),然後過程中兩數列使用等差級數和的公式直接去算有沒有相等。
下面是等差數列和的公式。
其中 a 是初始值、d 是公差、n 是數列的長度。
下面直接來看程式。
C++ 範例程式碼
class Solution {
public:
int sumArithmeticSeq(int startN, int endN) {
int n = endN - startN + 1;
return startN * n + (n * (n - 1) * 1) / 2;
}
int pivotInteger(int n) {
for (int i=n; i>0; --i) {
int a = sumArithmeticSeq(1, i);
int b = sumArithmeticSeq(i, n);
// If a < b, there is no answer
if (a < b) {
break;
}
if (a == b) {
return i;
}
}
return -1;
}
};
Python 範例程式碼
class Solution:
def sumArithmeticSeq(self, start_n: int, end_n: int) -> int:
n = end_n - start_n + 1
return start_n * n + (n * (n - 1) * 1) // 2;
def pivotInteger(self, n: int) -> int:
for i in reversed(range(1, n+1)):
a = self.sumArithmeticSeq(1, i)
b = self.sumArithmeticSeq(i, n)
# If a < b, there is no answer
if a < b:
return -1
elif a == b:
return i