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_1534_Count_Good_Triplets.java
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_1534_Count_Good_Triplets.java
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package com.github.junyu.solution.leetCode.easy.array;
public class _1534_Count_Good_Triplets {
/*Given an array of integers arr, and three integers a, b and c. You need to find the number of good triplets.
A triplet (arr[i], arr[j], arr[k]) is good if the following conditions are true:
0 <= i < j < k < arr.length
|arr[i] - arr[j]| <= a
|arr[j] - arr[k]| <= b
|arr[i] - arr[k]| <= c
Where |x| denotes the absolute value of x.
Return the number of good triplets.
Example 1:
Input: arr = [3,0,1,1,9,7], a = 7, b = 2, c = 3
Output: 4
Explanation: There are 4 good triplets: [(3,0,1), (3,0,1), (3,1,1), (0,1,1)].
Example 2:
Input: arr = [1,1,2,2,3], a = 0, b = 0, c = 1
Output: 0
Explanation: No triplet satisfies all conditions.
Constraints:
3 <= arr.length <= 100
0 <= arr[i] <= 1000
0 <= a, b, c <= 1000*/
/**
* 统计好三元组
* 思路:就是穷举所有的可能性,这里有一点值得注意 :&& 符号,如果左侧的结果为false,就不会再执行右侧的逻辑判断
* @param arr
* @param a
* @param b
* @param c
* @return
*/
public int countGoodTriplets(int[] arr, int a, int b, int c) {
int ans = 0;
for (int i = 0; i < arr.length - 2; i++) {
for (int j = i+1; j < arr.length - 1; j++) {
for (int k = j+1; k < arr.length; k++) {
if (Math.abs(arr[i] - arr[j]) <= a && Math.abs(arr[j] - arr[k]) <= b && Math.abs(arr[i] - arr[k]) <= c)
ans++;
}
}
}
return ans;
}
public static void main(String[] args) {
_1534_Count_Good_Triplets test = new _1534_Count_Good_Triplets();
System.out.println(test.countGoodTriplets(new int[]{3, 0, 1, 1, 9, 7}, 7, 2, 3));
System.out.println(test.countGoodTriplets(new int[]{1, 1, 2, 2, 3}, 0, 0, 1));
}
}