4545
4646## 解法
4747
48- ### 方法一
48+ ### 方法一:哈希表 + 枚举
49+ 50+ 我们用一个哈希表 $cnt$ 记录数组 $nums$ 中每个元素出现的次数。对于每个元素 $x,ドル我们可以将其不断平方,直到其值在哈希表 $cnt$ 中的出现次数小于 2ドル$ 为止。此时,我们判断 $x$ 在哈希表 $cnt$ 中的出现次数是否为 1ドル,ドル如果是则说明 $x$ 仍然可以被选入子集中,否则我们需要从子集中删除一个元素,确保子集个数为奇数。然后我们更新答案。继续枚举下一个元素。
51+ 52+ 注意我们需要特殊处理 $x = 1$ 的情况。
53+ 54+ 时间复杂度 $O(n \times \log \log M),ドル空间复杂度 $O(n)$。其中 $n$ 和 $M$ 分别是数组 $nums$ 的长度和数组 $nums$ 中的最大值。
4955
5056<!-- tabs:start -->
5157
5258``` python
5359class Solution :
5460 def maximumLength (self nums : List[int ]) -> int :
55- d = {}
56- for num in sorted (nums)[::- 1 ]:
57- if num** 2 in d and num in d and num != 1 :
58- d[num] = d[num** 2 ] + 2
59- else :
60- d[num] = 1
61- ones = nums.count(1 )
62- return max (max (d.values()), ones - (ones % 2 == 0 ))
63- 61+ cnt = Counter(nums)
62+ ans = cnt[1 ] - (cnt[1 ] % 2 ^ 1 )
63+ del cnt[1 ]
64+ for x in cnt:
65+ t = 0
66+ while cnt[x] > 1 :
67+ x = x * x
68+ t += 2
69+ t += 1 if cnt[x] else - 1
70+ ans = max (ans, t)
71+ return ans
6472```
6573
6674``` java
6775class Solution {
6876 public int maximumLength (int [] nums ) {
69- TreeMap< Integer , Integer >map = new TreeMap <>
70- for (int i : nums) {
71- map . put(i, map . getOrDefault(i, 0 ) + 1 );
77+ Map< Long , Integer >cnt = new HashMap <>
78+ for (int x : nums) {
79+ cnt . merge(( long ) x, 1 , Integer :: sum );
7280 }
73- int max = 0 ;
74- 75- for (Map . Entry<Integer , Integer > i : map. entrySet()) {
76- System . out. println(i. getValue());
77- if (i. getValue() >= 2 && i. getKey() != 1 ) {
78- int x = i. getKey();
79- int c = 2 ;
80- while (map. containsKey(x * x)) {
81- if (map. get(x * x) == 1 ) {
82- max = Math . max(max, c + 1 );
83- break ;
84- } else if (map. get(x * x) >= 2 ) {
85- max = Math . max(max, c + 1 );
86- x = x * x;
87- }
88- c += 2 ;
89- }
81+ Integer t = cnt. remove(1L );
82+ int ans = t == null ? 0 : t - (t % 2 ^ 1 );
83+ for (long x : cnt. keySet()) {
84+ t = 0 ;
85+ while (cnt. getOrDefault(x, 0 ) > 1 ) {
86+ x = x * x;
87+ t += 2 ;
9088 }
89+ t += cnt. getOrDefault(x, - 1 );
90+ ans = Math . max(ans, t);
9191 }
92- if (map. containsKey(1 ) && map. get(1 ) - 1 > max) {
93- return (map. get(1 ) % 2 != 0 ) ? map. get(1 ) : map. get(1 ) - 1 ;
94- }
95- return max == 0 ? 1 : max;
92+ return ans;
9693 }
9794}
9895```
@@ -101,90 +98,70 @@ class Solution {
10198class Solution {
10299public:
103100 int maximumLength(vector<int >& nums) {
104- long long ans = 0;
105- map<int, int> freq;
106- for (auto n : nums) {
107- freq[ n] ++;
101+ unordered_map<long long, int> cnt;
102+ for (int x : nums) {
103+ ++cnt[ x] ;
108104 }
109- for (auto [ k, f] : freq) {
110- long long t = k, count = 0;
111- if (t == 1) {
112- count += freq[ t] ;
113- freq[ t] = 0;
105+ int ans = cnt[ 1] - (cnt[ 1] % 2 ^ 1);
106+ cnt.erase(1);
107+ for (auto [ v, _ ] : cnt) {
108+ int t = 0;
109+ long long x = v;
110+ while (cnt.count(x) && cnt[ x] > 1) {
111+ x = x * x;
112+ t += 2;
114113 }
115- while (t < INT_MAX && freq[ t] > 0) {
116- count += 2;
117- if (freq[ t] == 1) {
118- break;
119- }
120- freq[ t] = 0;
121- t = t * t;
122- }
123- if (count % 2 == 0) {
124- count--;
125- }
126- ans = max(ans, count);
114+ t += cnt.count(x) ? 1 : -1;
115+ ans = max(ans, t);
127116 }
128117 return ans;
129118 }
130119};
131120```
132121
133122```go
134- func minExp(x, c int) (int, int) {
135- d := math.Sqrt(float64(x))
136- if d < 2 || float64(int(d)) < d {
137- return x, c
138- }
139- return minExp(int(d), c+1)
140- }
141- func maximumLength(nums []int) int {
142- m := make(map[int][]int)
143- for i := range nums {
144- base, c := minExp(nums[i], 1)
145- m[base] = append(m[base], c)
146- }
147- max := 1
148- for _, v := range m {
149- v := matchPattern(v)
150- max = Max(max, v)
123+ func maximumLength(nums []int) (ans int) {
124+ cnt := map[int]int{}
125+ for _, x := range nums {
126+ cnt[x]++
151127 }
152- _, ok := m[1]
153- if ok {
154- if len(m[1])%2 > 0 {
155- max = Max(max, len(m[1]))
128+ ans = cnt[1] - (cnt[1]%2 ^ 1)
129+ delete(cnt, 1)
130+ for x := range cnt {
131+ t := 0
132+ for cnt[x] > 1 {
133+ x = x * x
134+ t += 2
135+ }
136+ if cnt[x] > 0 {
137+ t += 1
156138 } else {
157- max = Max(max, len(m[1])-1)
139+ t -= 1
158140 }
141+ ans = max(ans, t)
159142 }
160- return max
143+ return
161144}
162- func Max(i, j int) int {
163- if i > j {
164- return i
165- }
166- return j
167- }
168- func matchPattern(arr []int) int {
169- sort.Slice(arr, func(i, j int) bool { return arr[i] < arr[j] })
170- start := arr[0]
171- bin := 2
172- for i := range arr {
173- if bin == 0 {
174- start++
175- bin = 2
176- }
177- if arr[i] == start {
178- bin--
179- }
180- }
181- if bin == 1 {
182- return 2*(start-arr[0]) + 1
183- } else if bin == 0 {
184- return 2*(start-arr[0]) + 1
185- } else {
186- return 2*(start-arr[0]) - 1
187- }
145+ ```
146+ 147+ ``` ts
148+ function maximumLength(nums : number []): number {
149+ const : Map <number , number > = new Map ();
150+ for (const of nums ) {
151+ cnt .set (x , (cnt .get (x ) ?? 0 ) + 1 );
152+ }
153+ let ans = cnt .has (1 ) ? cnt .get (1 )! - (cnt .get (1 )! % 2 ^ 1 ) : 0 ;
154+ cnt .delete (1 );
155+ for (let [x, _] of cnt ) {
156+ let t = 0 ;
157+ while (cnt .has (x ) && cnt .get (x )! > 1 ) {
158+ x = x * x ;
159+ t += 2 ;
160+ }
161+ t += cnt .has (x ) ? 1 : - 1 ;
162+ ans = Math .max (ans , t );
163+ }
164+ return ans ;
188165}
189166```
190167
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