文章目录
problem
413. Arithmetic Slices
An integer array is called arithmetic if it consists of at least three elements and if the difference between any two consecutive elements is the same.
For example, [1,3,5,7,9], [7,7,7,7], and [3,-1,-5,-9] are arithmetic sequences.
Given an integer array nums, return the number of arithmetic subarrays of nums.
A subarray is a contiguous subsequence of the array.
Example 1:
Input: nums = [1,2,3,4]
Output: 3
Explanation: We have 3 arithmetic slices in nums: [1, 2, 3], [2, 3, 4] and [1,2,3,4] itself.
Example 2:
Input: nums = [1]
Output: 0
approach 1 brute force
class Solution {
public:
int numberOfArithmeticSlices(vector<int>& nums) {
if(nums.size() < 3)return 0;
vector<int> diff(nums.size());
diff[0] = -114514;
for(int i=1; i<nums.size(); i++)
diff[i] = nums[i]-nums[i-1];
int res = 0, i=2;
while(i < nums.size()){
int cnt = 1;
while(i < nums.size() && diff[i-1]==diff[i])
++cnt, ++i;
if(cnt > 1) res += (cnt*(cnt-1))/2;
else ++i;
}
return res;
}
};
approach 2 without extra space
class Solution {
public:
int numberOfArithmeticSlices(vector<int>& nums) {
if(nums.size() < 3)return 0;
int res = 0, cnt = 1;
for(int i=2; i<nums.size(); i++){
if(nums[i]+nums[i-2] == 2*nums[i-1]){
++cnt;
}else{
res += (cnt*(cnt-1))/2;
cnt = 1;
}
}
if(cnt > 1) res +=(cnt*(cnt-1))/2;
return res;
}
};
approach 3 shortest code, but not best
class Solution {
public:
int numberOfArithmeticSlices(vector<int>& nums) {
if(nums.size() < 3)return 0;
int res = 0;
for(int i=2, j=1; i<nums.size(); i++){
if(nums[i]+nums[i-2] != 2*nums[i-1])
j=i;
res += i-j;
}
return res;
}
};
