2367. 算术三元组的数目
给你一个下标从 0 开始、严格递增 的整数数组 nums 和一个正整数 diff 。如果满足下述全部条件,则三元组 (i, j, k) 就是一个 算术三元组 :
i < j < k ,
nums[j] - nums[i] == diff 且
nums[k] - nums[j] == diff
返回不同 算术三元组 的数目。
示例 1:
输入:nums = [0,1,4,6,7,10], diff = 3
输出:2
解释:
(1, 2, 4) 是算术三元组:7 - 4 == 3 且 4 - 1 == 3 。
(2, 4, 5) 是算术三元组:10 - 7 == 3 且 7 - 4 == 3 。
示例 2:
输入:nums = [4,5,6,7,8,9], diff = 2
输出:2
解释:
(0, 2, 4) 是算术三元组:8 - 6 == 2 且 6 - 4 == 2 。
(1, 3, 5) 是算术三元组:9 - 7 == 2 且 7 - 5 == 2 。
提示:
3 <= nums.length <= 200
0 <= nums[i] <= 200
1 <= diff <= 50
nums 严格 递增
解答:直接三层循环暴力求解
class Solution {
public int arithmeticTriplets(int[] nums, int diff) {
int n=nums.length;
int res=0;
for(int i=0;i<n;i++){
for(int j=i+1;j<n;j++){
if(nums[j]-nums[i]>diff)continue;
for(int k=j+1;k<n;k++){
if(nums[k]-nums[j]>diff)continue;
if(nums[j]-nums[i]==diff&&nums[k]-nums[j]==diff) {
res++;
}
}
}
}
return res;
}
}