Java C++题解leetcode判定是否为字符重排
更新时间:2022年09月29日 15:01:51 作者:AnjaVon
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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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