Java C++题解leetcode字符串轮转KMP算法详解
作者:AnjaVon
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题目要求
思路一:双指针(模拟)
Java
class Solution { public boolean isFlipedString(String s1, String s2) { if (s1.length() != s2.length()) return false; int n = s1.length(); if (n == 0) return true; for (int i = 0; i < n; i++) { boolean res = true; for (int j = 0; j < n; j++) { if (s1.charAt((i + j) % n) != s2.charAt(j)) { res = false; break; } } if (res) return true; } return false; } }
- 时间复杂度:O(n^2)
- 空间复杂度:O(1)
C++
class Solution { public: bool isFlipedString(string s1, string s2) { if (s1.size() != s2.size()) return false; int n = s1.size(); if (n == 0) return true; for (int i = 0; i < n; i++) { bool res = true; for (int j = 0; j < n; j++) { if (s1[(i + j) % n] != s2[j]) { res = false; break; } } if (res) return true; } return false; } };
- 时间复杂度:O(n^2)
- 空间复杂度:O(1)
思路二:子串
手写KMP
Java
get_next
class Solution { public boolean isFlipedString(String s1, String s2) { if (s1.length() != s2.length()) return false; int n = s1.length(); if (n == 0) return true; int[] nxt = new int[n]; nxt[0] = -1; int j = 0; // pat指针 int k = -1; // 跳转位置 while (j < n - 1) { if (k == -1 || s2.charAt(j) == s2.charAt(k)) { if (s2.charAt(++j) == s2.charAt(++k)) nxt[j] = nxt[k]; // 连续相同 else nxt[j] = k; } else k = nxt[k]; } String txt = s1 + s1; j = 0; for (int i = 0; i < 2 * n; i++) { if (j < n && txt.charAt(i) == s2.charAt(j)) j++; else { j = nxt[j]; if (j == -1) j++; } if (j == n) return true; } return false; } }
dp
class Solution { public boolean isFlipedString(String s1, String s2) { if (s1.length() != s2.length()) return false; int n = s1.length(); if (n == 0) return true; int[][] dp = new int[n][256]; // dp[state][char] = nxt state dp[0][s2.charAt(0)] = 1; int x = 0; // 影子状态 for (int s = 1; s < n; s++) { for (int c = 0; c < 256; c++) dp[s][c] = dp[x][c]; dp[s][s2.charAt(s)] = s + 1; x = dp[x][s2.charAt(s)]; } String txt = s1 + s1; int state = 0; for (int i = 0; i < 2 * n; i++) { state = dp[state][txt.charAt(i)]; if (state == n) return true; } return false; } }
- 时间复杂度:O(n)
- 空间复杂度:O(n)
C++
get_next
class Solution { public: bool isFlipedString(string s1, string s2) { if (s1.size() != s2.size()) return false; int n = s1.size(); if (n == 0) return true; int nxt[n]; nxt[0] = -1; int j = 0; // pat指针 int k = -1; // 跳转位置 while (j < n - 1) { if (k == -1 || s2[j] == s2[k]) { if (s2[++j] == s2[++k]) nxt[j] = nxt[k]; // 连续相同 else nxt[j] = k; } else k = nxt[k]; } string txt = s1 + s1; j = 0; for (int i = 0; i < 2 * n; i++) { if (j < n && txt[i] == s2[j]) j++; else { j = nxt[j]; if (j == -1) j++; } if (j == n) return true; } return false; } };
dp
class Solution { public: bool isFlipedString(string s1, string s2) { if (s1.size() != s2.size()) return false; int n = s1.size(); if (n == 0) return true; int dp[n][256]; // dp[state][char] = nxt state memset(dp, 0, sizeof(dp)); dp[0][s2[0]] = 1; int x = 0; // 影子状态 for (int s = 1; s < n; s++) { for (int c = 0; c < 256; c++) dp[s][c] = dp[x][c]; dp[s][s2[s]] = s + 1; x = dp[x][s2[s]]; } string txt = s1 + s1; int state = 0; for (int i = 0; i < 2 * n; i++) { state = dp[state][txt[i]]; if (state == n) return true; } return false; } };
- 时间复杂度:O(n)
- 空间复杂度:O(n)
调API
Java
class Solution { public boolean isFlipedString(String s1, String s2) { return s1.length() == s2.length() && (s1 + s1).contains(s2); } }
- 时间复杂度:O(n),取决于语言匹配子字符串机制
- 空间复杂度:O(n)
C++
class Solution { public: bool isFlipedString(string s1, string s2) { return s1.size() == s2.size() && (s1 + s1).find(s2) != string::npos; } };
- 时间复杂度:O(n),取决于语言匹配子字符串机制
- 空间复杂度:O(n)
impl Solution { pub fn is_fliped_string(s1: String, s2: String) -> bool { s1.len() == s2.len() && s2.repeat(2).contains(&s1) } }
- 时间复杂度:O(n),取决于语言匹配子字符串机制
- 空间复杂度:O(n)
总结
做过了轮转的题就能很快意识到拼接然后找子串,本来调个API结束结果发现了KMP算法,就浅学一下~
时间耗在算法了就掠过rust
各种辣~
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