题目要求
思路一：排序
Java
class Solution {
public boolean CheckPermutation(String s1, String s2) {
if(s1.length() != s2.length())
return false;
char[] sort1 = s1.toCharArray();
Arrays.sort(sort1);
char[] sort2 = s2.toCharArray();
Arrays.sort(sort2);
return Arrays.equals(sort1, sort2);
}
}
- 时间复杂度：O(n log n)，排序复杂度
- 空间复杂度：O(n)，拷贝字符串用于排序
C++
class Solution {
public:
bool CheckPermutation(string s1, string s2) {
if (s1.size() != s2.size())
return false;
sort(s1.begin(), s1.end());
sort(s2.begin(), s2.end());
return s1 == s2;
}
};
- 时间复杂度：O(nlogn)O(n\log n)O(nlogn)，排序复杂度
- 空间复杂度：O(logn)O(\log n)O(logn)，排序需要
Rust
impl Solution {
pub fn check_permutation(s1: String, s2: String) -> bool {
if s1.len() != s2.len() {
false
}
else {
let (mut sort1, mut sort2) = (s1.as_bytes().to_vec(), s2.as_bytes().to_vec());
sort1.sort();
sort2.sort();
sort1 == sort2
}
}
}
- 时间复杂度：O(n log n)，排序复杂度
- 空间复杂度：O(n)，拷贝字符串用于排序
思路二：词频统计
Java
class Solution {
public boolean CheckPermutation(String s1, String s2) {
if(s1.length() != s2.length())
return false;
int[] freq = new int[128];
int diff = 0;
for (int i = 0; i < s1.length(); i++) {
if (++freq[s1.charAt(i)] == 1)
diff++;
if (--freq[s2.charAt(i)] == 0)
diff--;
}
return diff == 0;
}
}
- 时间复杂度：O(n)
- 空间复杂度：O(C)，常数C为字符集大小
C++
class Solution {
public:
bool CheckPermutation(string s1, string s2) {
if (s1.size() != s2.size())
return false;
int freq[128];
memset(freq, 0, sizeof(freq));
int diff = 0;
for (int i = 0; i < s1.size(); i++) {
if (++freq[s1[i]] == 1)
diff++;
if (--freq[s2[i]] == 0)
diff--;
}
return diff == 0;
}
};
- 时间复杂度：O(n)
- 空间复杂度：O(C)，常数C为字符集大小
Rust
impl Solution {
pub fn check_permutation(s1: String, s2: String) -> bool {
s1.len() == s2.len() && s1.bytes().zip(s2.bytes()).fold(vec![0; 128], |mut freq, (c1, c2)| {
freq[c1 as usize] += 1;
freq[c2 as usize] -= 1;
freq
}).into_iter().all(|diff| diff == 0)
}
}
- 时间复杂度：O(n)
- 空间复杂度：O(C)，常数C为字符集大小
总结
简单模拟题、快乐结束~
有些语言不能改的字符串在这种时候真是烦烦……
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