| comments | difficulty | edit_url | rating | source | tags | ||||
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true |
中等 |
1591 |
第 15 场双周赛 Q3 |
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请你设计一个迭代器类 CombinationIterator ，包括以下内容：
CombinationIterator(string characters, int combinationLength)一个构造函数，输入参数包括：用一个 有序且字符唯一 的字符串characters（该字符串只包含小写英文字母）和一个数字combinationLength。- 函数
next()，按 字典序 返回长度为combinationLength的下一个字母组合。 - 函数
hasNext()，只有存在长度为combinationLength的下一个字母组合时，才返回true
示例 1：
输入:
["CombinationIterator", "next", "hasNext", "next", "hasNext", "next", "hasNext"]
[["abc", 2], [], [], [], [], [], []]
输出：
[null, "ab", true, "ac", true, "bc", false]
解释：
CombinationIterator iterator = new CombinationIterator("abc", 2); // 创建迭代器 iterator
iterator.next(); // 返回 "ab"
iterator.hasNext(); // 返回 true
iterator.next(); // 返回 "ac"
iterator.hasNext(); // 返回 true
iterator.next(); // 返回 "bc"
iterator.hasNext(); // 返回 false
提示：
1 <= combinationLength <= characters.length <= 15-
characters中每个字符都 不同 - 每组测试数据最多对
next和hasNext调用104次 - 题目保证每次调用函数
next时都存在下一个字母组合。
我们通过
class CombinationIterator:
def __init__(self, characters: str, combinationLength: int):
def dfs(i):
if len(t) == combinationLength:
cs.append(''.join(t))
return
if i == n:
return
t.append(characters[i])
dfs(i + 1)
t.pop()
dfs(i + 1)
cs = []
n = len(characters)
t = []
dfs(0)
self.cs = cs
self.idx = 0
def next(self) -> str:
ans = self.cs[self.idx]
self.idx += 1
return ans
def hasNext(self) -> bool:
return self.idx < len(self.cs)
# Your CombinationIterator object will be instantiated and called as such:
# obj = CombinationIterator(characters, combinationLength)
# param_1 = obj.next()
# param_2 = obj.hasNext()class CombinationIterator {
private int n;
private int combinationLength;
private String characters;
private StringBuilder t = new StringBuilder();
private List<String> cs = new ArrayList<>();
private int idx = 0;
public CombinationIterator(String characters, int combinationLength) {
n = characters.length();
this.combinationLength = combinationLength;
this.characters = characters;
dfs(0);
}
public String next() {
return cs.get(idx++);
}
public boolean hasNext() {
return idx < cs.size();
}
private void dfs(int i) {
if (t.length() == combinationLength) {
cs.add(t.toString());
return;
}
if (i == n) {
return;
}
t.append(characters.charAt(i));
dfs(i + 1);
t.deleteCharAt(t.length() - 1);
dfs(i + 1);
}
}
/**
* Your CombinationIterator object will be instantiated and called as such:
* CombinationIterator obj = new CombinationIterator(characters, combinationLength);
* String param_1 = obj.next();
* boolean param_2 = obj.hasNext();
*/class CombinationIterator {
public:
string characters;
vector<string> cs;
int idx;
int n;
int combinationLength;
string t;
CombinationIterator(string characters, int combinationLength) {
idx = 0;
n = characters.size();
this->characters = characters;
this->combinationLength = combinationLength;
dfs(0);
}
string next() {
return cs[idx++];
}
bool hasNext() {
return idx < cs.size();
}
void dfs(int i) {
if (t.size() == combinationLength) {
cs.push_back(t);
return;
}
if (i == n) return;
t.push_back(characters[i]);
dfs(i + 1);
t.pop_back();
dfs(i + 1);
}
};
/**
* Your CombinationIterator object will be instantiated and called as such:
* CombinationIterator* obj = new CombinationIterator(characters, combinationLength);
* string param_1 = obj->next();
* bool param_2 = obj->hasNext();
*/type CombinationIterator struct {
cs []string
idx int
}
func Constructor(characters string, combinationLength int) CombinationIterator {
t := []byte{}
n := len(characters)
cs := []string{}
var dfs func(int)
dfs = func(i int) {
if len(t) == combinationLength {
cs = append(cs, string(t))
return
}
if i == n {
return
}
t = append(t, characters[i])
dfs(i + 1)
t = t[:len(t)-1]
dfs(i + 1)
}
dfs(0)
return CombinationIterator{cs, 0}
}
func (this *CombinationIterator) Next() string {
ans := this.cs[this.idx]
this.idx++
return ans
}
func (this *CombinationIterator) HasNext() bool {
return this.idx < len(this.cs)
}
/**
* Your CombinationIterator object will be instantiated and called as such:
* obj := Constructor(characters, combinationLength);
* param_1 := obj.Next();
* param_2 := obj.HasNext();
*/我们看个例子，对于
对应的二进制数为：
1100
1010
1001
0110
0101
0011
...
观察到上述规律后，我们依次按照二进制编码从大到小的规律，将所有字符串依次求出。
所谓的长度
class CombinationIterator:
def __init__(self, characters: str, combinationLength: int):
self.curr = (1 << len(characters)) - 1
self.size = combinationLength
self.cs = characters[::-1]
def next(self) -> str:
while self.curr >= 0 and self.curr.bit_count() != self.size:
self.curr -= 1
ans = []
for i in range(len(self.cs)):
if (self.curr >> i) & 1:
ans.append(self.cs[i])
self.curr -= 1
return ''.join(ans[::-1])
def hasNext(self) -> bool:
while self.curr >= 0 and self.curr.bit_count() != self.size:
self.curr -= 1
return self.curr >= 0
# Your CombinationIterator object will be instantiated and called as such:
# obj = CombinationIterator(characters, combinationLength)
# param_1 = obj.next()
# param_2 = obj.hasNext()class CombinationIterator {
private int curr;
private int size;
private char[] cs;
public CombinationIterator(String characters, int combinationLength) {
int n = characters.length();
curr = (1 << n) - 1;
size = combinationLength;
cs = new char[n];
for (int i = 0; i < n; ++i) {
cs[i] = characters.charAt(n - i - 1);
}
}
public String next() {
while (curr >= 0 && Integer.bitCount(curr) != size) {
--curr;
}
StringBuilder ans = new StringBuilder();
for (int i = 0; i < cs.length; ++i) {
if (((curr >> i) & 1) == 1) {
ans.append(cs[i]);
}
}
--curr;
return ans.reverse().toString();
}
public boolean hasNext() {
while (curr >= 0 && Integer.bitCount(curr) != size) {
--curr;
}
return curr >= 0;
}
}
/**
* Your CombinationIterator object will be instantiated and called as such:
* CombinationIterator obj = new CombinationIterator(characters, combinationLength);
* String param_1 = obj.next();
* boolean param_2 = obj.hasNext();
*/class CombinationIterator {
public:
int size;
string cs;
int curr;
CombinationIterator(string characters, int combinationLength) {
int n = characters.size();
curr = (1 << n) - 1;
reverse(characters.begin(), characters.end());
cs = characters;
size = combinationLength;
}
string next() {
while (curr >= 0 && __builtin_popcount(curr) != size) --curr;
string ans;
for (int i = 0; i < cs.size(); ++i) {
if ((curr >> i) & 1) {
ans += cs[i];
}
}
reverse(ans.begin(), ans.end());
--curr;
return ans;
}
bool hasNext() {
while (curr >= 0 && __builtin_popcount(curr) != size) --curr;
return curr >= 0;
}
};
/**
* Your CombinationIterator object will be instantiated and called as such:
* CombinationIterator* obj = new CombinationIterator(characters, combinationLength);
* string param_1 = obj->next();
* bool param_2 = obj->hasNext();
*/type CombinationIterator struct {
curr int
size int
cs []byte
}
func Constructor(characters string, combinationLength int) CombinationIterator {
n := len(characters)
curr := (1 << n) - 1
size := combinationLength
cs := make([]byte, n)
for i := range characters {
cs[n-i-1] = characters[i]
}
return CombinationIterator{curr, size, cs}
}
func (this *CombinationIterator) Next() string {
for this.curr >= 0 && bits.OnesCount(uint(this.curr)) != this.size {
this.curr--
}
ans := []byte{}
for i := range this.cs {
if (this.curr >> i & 1) == 1 {
ans = append(ans, this.cs[i])
}
}
for i, j := 0, len(ans)-1; i < j; i, j = i+1, j-1 {
ans[i], ans[j] = ans[j], ans[i]
}
this.curr--
return string(ans)
}
func (this *CombinationIterator) HasNext() bool {
for this.curr >= 0 && bits.OnesCount(uint(this.curr)) != this.size {
this.curr--
}
return this.curr >= 0
}
/**
* Your CombinationIterator object will be instantiated and called as such:
* obj := Constructor(characters, combinationLength);
* param_1 := obj.Next();
* param_2 := obj.HasNext();
*/