给你一个由 正 整数组成的数组 nums 。
如果数组中的某个子数组满足下述条件，则称之为 完全子数组 ：
- 子数组中 不同 元素的数目等于整个数组不同元素的数目。
返回数组中 完全子数组 的数目。
子数组 是数组中的一个连续非空序列。
示例 1：
输入：nums = [1,3,1,2,2] 输出：4 解释：完全子数组有：[1,3,1,2]、[1,3,1,2,2]、[3,1,2] 和 [3,1,2,2] 。
示例 2：
输入：nums = [5,5,5,5] 输出：10 解释：数组仅由整数 5 组成，所以任意子数组都满足完全子数组的条件。子数组的总数为 10 。
提示：
1 <= nums.length <= 10001 <= nums[i] <= 2000
方法一：哈希表 + 枚举
我们先用哈希表统计数组中不同元素的数目，记为
接下来，我们枚举子数组的左端点下标
枚举结束后，返回答案即可。
时间复杂度
方法二：哈希表 + 双指针
与方法一类似，我们可以使用哈希表统计数组中不同元素的数目，记为
接下来，我们使用双指针维护一个滑动窗口，滑动窗口的右端点下标为
每次固定左端点下标
时间复杂度
class Solution:
def countCompleteSubarrays(self, nums: List[int]) -> int:
cnt = len(set(nums))
ans, n = 0, len(nums)
for i in range(n):
s = set()
for x in nums[i:]:
s.add(x)
if len(s) == cnt:
ans += 1
return ansclass Solution:
def countCompleteSubarrays(self, nums: List[int]) -> int:
cnt = len(set(nums))
d = Counter()
ans, n = 0, len(nums)
i = 0
for j, x in enumerate(nums):
d[x] += 1
while len(d) == cnt:
ans += n - j
d[nums[i]] -= 1
if d[nums[i]] == 0:
d.pop(nums[i])
i += 1
return ansclass Solution {
public int countCompleteSubarrays(int[] nums) {
Set<Integer> s = new HashSet<>();
for (int x : nums) {
s.add(x);
}
int cnt = s.size();
int ans = 0, n = nums.length;
for (int i = 0; i < n; ++i) {
s.clear();
for (int j = i; j < n; ++j) {
s.add(nums[j]);
if (s.size() == cnt) {
++ans;
}
}
}
return ans;
}
}class Solution {
public int countCompleteSubarrays(int[] nums) {
Map<Integer, Integer> d = new HashMap<>();
for (int x : nums) {
d.put(x, 1);
}
int cnt = d.size();
int ans = 0, n = nums.length;
d.clear();
for (int i = 0, j = 0; j < n; ++j) {
d.merge(nums[j], 1, Integer::sum);
while (d.size() == cnt) {
ans += n - j;
if (d.merge(nums[i], -1, Integer::sum) == 0) {
d.remove(nums[i]);
}
++i;
}
}
return ans;
}
}class Solution {
public:
int countCompleteSubarrays(vector<int>& nums) {
unordered_set<int> s(nums.begin(), nums.end());
int cnt = s.size();
int ans = 0, n = nums.size();
for (int i = 0; i < n; ++i) {
s.clear();
for (int j = i; j < n; ++j) {
s.insert(nums[j]);
if (s.size() == cnt) {
++ans;
}
}
}
return ans;
}
};class Solution {
public:
int countCompleteSubarrays(vector<int>& nums) {
unordered_map<int, int> d;
for (int x : nums) {
d[x] = 1;
}
int cnt = d.size();
d.clear();
int ans = 0, n = nums.size();
for (int i = 0, j = 0; j < n; ++j) {
d[nums[j]]++;
while (d.size() == cnt) {
ans += n - j;
if (--d[nums[i]] == 0) {
d.erase(nums[i]);
}
++i;
}
}
return ans;
}
};func countCompleteSubarrays(nums []int) (ans int) {
s := map[int]bool{}
for _, x := range nums {
s[x] = true
}
cnt := len(s)
for i := range nums {
s = map[int]bool{}
for _, x := range nums[i:] {
s[x] = true
if len(s) == cnt {
ans++
}
}
}
return
}func countCompleteSubarrays(nums []int) (ans int) {
d := map[int]int{}
for _, x := range nums {
d[x] = 1
}
cnt := len(d)
i, n := 0, len(nums)
d = map[int]int{}
for j, x := range nums {
d[x]++
for len(d) == cnt {
ans += n - j
d[nums[i]]--
if d[nums[i]] == 0 {
delete(d, nums[i])
}
i++
}
}
return
}function countCompleteSubarrays(nums: number[]): number {
const s: Set<number> = new Set(nums);
const cnt = s.size;
const n = nums.length;
let ans = 0;
for (let i = 0; i < n; ++i) {
s.clear();
for (let j = i; j < n; ++j) {
s.add(nums[j]);
if (s.size === cnt) {
++ans;
}
}
}
return ans;
}function countCompleteSubarrays(nums: number[]): number {
const d: Map<number, number> = new Map();
for (const x of nums) {
d.set(x, (d.get(x) ?? 0) + 1);
}
const cnt = d.size;
d.clear();
const n = nums.length;
let ans = 0;
let i = 0;
for (let j = 0; j < n; ++j) {
d.set(nums[j], (d.get(nums[j]) ?? 0) + 1);
while (d.size === cnt) {
ans += n - j;
d.set(nums[i], d.get(nums[i])! - 1);
if (d.get(nums[i]) === 0) {
d.delete(nums[i]);
}
++i;
}
}
return ans;
}use std::collections::HashSet;
impl Solution {
pub fn count_complete_subarrays(nums: Vec<i32>) -> i32 {
let mut set: HashSet<&i32> = nums.iter().collect();
let n = nums.len();
let m = set.len();
let mut ans = 0;
for i in 0..n {
set.clear();
for j in i..n {
set.insert(&nums[j]);
if set.len() == m {
ans += n - j;
break;
}
}
}
ans as i32
}
}use std::collections::HashMap;
use std::collections::HashSet;
impl Solution {
pub fn count_complete_subarrays(nums: Vec<i32>) -> i32 {
let n = nums.len();
let m = nums.iter().collect::<HashSet<&i32>>().len();
let mut map = HashMap::new();
let mut ans = 0;
let mut i = 0;
for j in 0..n {
*map.entry(nums[j]).or_insert(0) += 1;
while map.len() == m {
ans += n - j;
let v = map.entry(nums[i]).or_default();
*v -= 1;
if *v == 0 {
map.remove(&nums[i]);
}
i += 1;
}
}
ans as i32
}
}