题目描述
给你一个 下标从 0 开始 的整数数组 nums ，返回满足下述条件的 不同 四元组 (a, b, c, d) 的 数目 ：
nums[a] + nums[b] + nums[c] == nums[d] ，且a < b < c < d
示例 1：
输入：nums = [1,2,3,6]
输出：1
解释：满足要求的唯一一个四元组是 (0, 1, 2, 3) 因为 1 + 2 + 3 == 6 。
示例 2：
输入：nums = [3,3,6,4,5]
输出：0
解释：[3,3,6,4,5] 中不存在满足要求的四元组。
示例 3：
输入：nums = [1,1,1,3,5]
输出：4
解释：满足要求的 4 个四元组如下：
- (0, 1, 2, 3): 1 + 1 + 1 == 3
- (0, 1, 3, 4): 1 + 1 + 3 == 5
- (0, 2, 3, 4): 1 + 1 + 3 == 5
- (1, 2, 3, 4): 1 + 1 + 3 == 5
提示：
4 <= nums.length <= 501 <= nums[i] <= 100
解法
方法一
| class Solution:
def countQuadruplets(self, nums: List[int]) -> int:
ans, n = 0, len(nums)
for a in range(n - 3):
for b in range(a + 1, n - 2):
for c in range(b + 1, n - 1):
for d in range(c + 1, n):
if nums[a] + nums[b] + nums[c] == nums[d]:
ans += 1
return ans
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17 | class Solution {
public int countQuadruplets(int[] nums) {
int ans = 0, n = nums.length;
for (int a = 0; a < n - 3; ++a) {
for (int b = a + 1; b < n - 2; ++b) {
for (int c = b + 1; c < n - 1; ++c) {
for (int d = c + 1; d < n; ++d) {
if (nums[a] + nums[b] + nums[c] == nums[d]) {
++ans;
}
}
}
}
}
return ans;
}
}
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12 | class Solution {
public:
int countQuadruplets(vector<int>& nums) {
int ans = 0, n = nums.size();
for (int a = 0; a < n - 3; ++a)
for (int b = a + 1; b < n - 2; ++b)
for (int c = b + 1; c < n - 1; ++c)
for (int d = c + 1; d < n; ++d)
if (nums[a] + nums[b] + nums[c] == nums[d]) ++ans;
return ans;
}
};
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15 | func countQuadruplets(nums []int) int {
ans, n := 0, len(nums)
for a := 0; a < n-3; a++ {
for b := a + 1; b < n-2; b++ {
for c := b + 1; c < n-1; c++ {
for d := c + 1; d < n; d++ {
if nums[a]+nums[b]+nums[c] == nums[d] {
ans++
}
}
}
}
}
return ans
}
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方法二
| class Solution:
def countQuadruplets(self, nums: List[int]) -> int:
ans, n = 0, len(nums)
counter = Counter()
for c in range(n - 2, 1, -1):
counter[nums[c + 1]] += 1
for a in range(c - 1):
for b in range(a + 1, c):
ans += counter[nums[a] + nums[b] + nums[c]]
return ans
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15 | class Solution {
public int countQuadruplets(int[] nums) {
int ans = 0, n = nums.length;
int[] counter = new int[310];
for (int c = n - 2; c > 1; --c) {
++counter[nums[c + 1]];
for (int a = 0; a < c - 1; ++a) {
for (int b = a + 1; b < c; ++b) {
ans += counter[nums[a] + nums[b] + nums[c]];
}
}
}
return ans;
}
}
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16 | class Solution {
public:
int countQuadruplets(vector<int>& nums) {
int ans = 0, n = nums.size();
vector<int> counter(310);
for (int c = n - 2; c > 1; --c) {
++counter[nums[c + 1]];
for (int a = 0; a < c - 1; ++a) {
for (int b = a + 1; b < c; ++b) {
ans += counter[nums[a] + nums[b] + nums[c]];
}
}
}
return ans;
}
};
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13 | func countQuadruplets(nums []int) int {
ans, n := 0, len(nums)
counter := make([]int, 310)
for c := n - 2; c > 1; c-- {
counter[nums[c+1]]++
for a := 0; a < c-1; a++ {
for b := a + 1; b < c; b++ {
ans += counter[nums[a]+nums[b]+nums[c]]
}
}
}
return ans
}
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方法三
| class Solution:
def countQuadruplets(self, nums: List[int]) -> int:
ans, n = 0, len(nums)
counter = Counter()
for b in range(n - 3, 0, -1):
c = b + 1
for d in range(c + 1, n):
counter[nums[d] - nums[c]] += 1
for a in range(b):
ans += counter[nums[a] + nums[b]]
return ans
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18 | class Solution {
public int countQuadruplets(int[] nums) {
int ans = 0, n = nums.length;
int[] counter = new int[310];
for (int b = n - 3; b > 0; --b) {
int c = b + 1;
for (int d = c + 1; d < n; ++d) {
if (nums[d] - nums[c] >= 0) {
++counter[nums[d] - nums[c]];
}
}
for (int a = 0; a < b; ++a) {
ans += counter[nums[a] + nums[b]];
}
}
return ans;
}
}
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19 | class Solution {
public:
int countQuadruplets(vector<int>& nums) {
int ans = 0, n = nums.size();
vector<int> counter(310);
for (int b = n - 3; b > 0; --b) {
int c = b + 1;
for (int d = c + 1; d < n; ++d) {
if (nums[d] - nums[c] >= 0) {
++counter[nums[d] - nums[c]];
}
}
for (int a = 0; a < b; ++a) {
ans += counter[nums[a] + nums[b]];
}
}
return ans;
}
};
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16 | func countQuadruplets(nums []int) int {
ans, n := 0, len(nums)
counter := make([]int, 310)
for b := n - 3; b > 0; b-- {
c := b + 1
for d := c + 1; d < n; d++ {
if nums[d] >= nums[c] {
counter[nums[d]-nums[c]]++
}
}
for a := 0; a < b; a++ {
ans += counter[nums[a]+nums[b]]
}
}
return ans
}
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