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Commit f1d743e

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feat: add solutions to lc problem: No.1542
No.1542.Find Longest Awesome Substring
1 parent a2d5c34 commit f1d743e

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6 files changed

+299
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lines changed

6 files changed

+299
-2
lines changed

‎solution/1500-1599/1542.Find Longest Awesome Substring/README.md‎

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@@ -56,22 +56,133 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:状态压缩 + 前缀和思想**
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根据题目描述,"超赞子字符串"中的字符可以通过交换得到回文字符串,因此,"超赞子字符串"中最多有一个数字字符出现奇数次,其余数字字符出现偶数次。
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我们可以用一个整数 $st$ 来表示当前前缀字符串中数字字符出现的奇偶性,其中 $st$ 的第 $i$ 位表示数字字符 $i$ 出现的奇偶性,即 $st$ 的第 $i$ 位为 1ドル$ 表示数字字符 $i$ 出现奇数次,为 0ドル$ 表示数字字符 $i$ 出现偶数次。
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而如果子字符串 $s[j,..i]$ 是"超赞字符串",那么前缀字符串 $s[0,..i]$ 的状态 $st$ 与前缀字符串 $s[0,..j-1]$ 的状态 $st'$ 的二进制位中,最多只有一位不同。这是因为,二进制位不同,表示奇偶性不同,而奇偶性不同,就意味着子字符串 $s[j,..i]$ 中该数字出现的次数为奇数次。
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所以,我们可以用哈希表或数组记录所有状态 $st$ 第一次出现的位置。若当前前缀字符串的状态 $st$ 在哈希表中已经存在,那么说明当前前缀字符串的状态 $st$ 与前缀字符串 $s[0,..j-1]$ 的状态 $st'$ 的二进制位中,所有位都相同,即子字符串 $s[j,..i]$ 是"超赞字符串",更新答案的最大值。或者,我们可以枚举每一位,将当前前缀字符串的状态 $st$ 的第 $i$ 位取反,即 $st \oplus (1 << i),ドル然后判断 $st \oplus (1 << i)$ 是否在哈希表中,若在,那么说明当前前缀字符串的状态 $st$ 与前缀字符串 $s[0,..j-1]$ 的状态 $st' \oplus (1 << i)$ 的二进制位中,只有第 $i$ 位不同,即子字符串 $s[j,..i]$ 是"超赞字符串",更新答案的最大值。
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最后,返回答案即可。
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时间复杂度 $O(n \times C),ドル空间复杂度 $O(2^C)$。其中 $n$ 和 $C$ 分别为字符串 $s$ 的长度和数字字符的种类数。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def longestAwesome(self, s: str) -> int:
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st = 0
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d = {0: -1}
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ans = 1
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for i, c in enumerate(s):
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v = int(c)
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st ^= 1 << v
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if st in d:
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ans = max(ans, i - d[st])
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else:
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d[st] = i
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for v in range(10):
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if st ^ (1 << v) in d:
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ans = max(ans, i - d[st ^ (1 << v)])
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int longestAwesome(String s) {
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int[] d = new int[1024];
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int st = 0, ans = 1;
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Arrays.fill(d, -1);
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d[0] = 0;
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for (int i = 1; i <= s.length(); ++i) {
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int v = s.charAt(i - 1) - '0';
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st ^= 1 << v;
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if (d[st] >= 0) {
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ans = Math.max(ans, i - d[st]);
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} else {
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d[st] = i;
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}
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for (v = 0; v < 10; ++v) {
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if (d[st ^ (1 << v)] >= 0) {
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ans = Math.max(ans, i - d[st ^ (1 << v)]);
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}
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int longestAwesome(string s) {
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vector<int> d(1024, -1);
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d[0] = 0;
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int st = 0, ans = 1;
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for (int i = 1; i <= s.size(); ++i) {
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int v = s[i - 1] - '0';
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st ^= 1 << v;
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if (~d[st]) {
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ans = max(ans, i - d[st]);
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} else {
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d[st] = i;
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}
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for (v = 0; v < 10; ++v) {
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if (~d[st ^ (1 << v)]) {
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ans = max(ans, i - d[st ^ (1 << v)]);
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}
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func longestAwesome(s string) int {
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d := [1024]int{}
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d[0] = 1
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st, ans := 0, 1
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for i, c := range s {
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i += 2
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st ^= 1 << (c - '0')
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if d[st] > 0 {
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ans = max(ans, i-d[st])
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} else {
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d[st] = i
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}
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for v := 0; v < 10; v++ {
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if d[st^(1<<v)] > 0 {
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ans = max(ans, i-d[st^(1<<v)])
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}
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}
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}
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return ans
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**

‎solution/1500-1599/1542.Find Longest Awesome Substring/README_EN.md‎

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### **Python3**
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```python
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class Solution:
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def longestAwesome(self, s: str) -> int:
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st = 0
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d = {0: -1}
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ans = 1
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for i, c in enumerate(s):
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v = int(c)
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st ^= 1 << v
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if st in d:
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ans = max(ans, i - d[st])
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else:
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d[st] = i
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for v in range(10):
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if st ^ (1 << v) in d:
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ans = max(ans, i - d[st ^ (1 << v)])
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return ans
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```
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### **Java**
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```java
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class Solution {
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public int longestAwesome(String s) {
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int[] d = new int[1024];
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int st = 0, ans = 1;
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Arrays.fill(d, -1);
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d[0] = 0;
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for (int i = 1; i <= s.length(); ++i) {
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int v = s.charAt(i - 1) - '0';
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st ^= 1 << v;
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if (d[st] >= 0) {
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ans = Math.max(ans, i - d[st]);
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} else {
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d[st] = i;
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}
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for (v = 0; v < 10; ++v) {
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if (d[st ^ (1 << v)] >= 0) {
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ans = Math.max(ans, i - d[st ^ (1 << v)]);
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}
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int longestAwesome(string s) {
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vector<int> d(1024, -1);
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d[0] = 0;
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int st = 0, ans = 1;
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for (int i = 1; i <= s.size(); ++i) {
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int v = s[i - 1] - '0';
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st ^= 1 << v;
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if (~d[st]) {
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ans = max(ans, i - d[st]);
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} else {
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d[st] = i;
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}
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for (v = 0; v < 10; ++v) {
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if (~d[st ^ (1 << v)]) {
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ans = max(ans, i - d[st ^ (1 << v)]);
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}
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func longestAwesome(s string) int {
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d := [1024]int{}
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d[0] = 1
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st, ans := 0, 1
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for i, c := range s {
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i += 2
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st ^= 1 << (c - '0')
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if d[st] > 0 {
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ans = max(ans, i-d[st])
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} else {
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d[st] = i
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}
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for v := 0; v < 10; v++ {
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if d[st^(1<<v)] > 0 {
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ans = max(ans, i-d[st^(1<<v)])
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}
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}
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}
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return ans
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**
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class Solution {
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public:
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int longestAwesome(string s) {
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vector<int> d(1024, -1);
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d[0] = 0;
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int st = 0, ans = 1;
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for (int i = 1; i <= s.size(); ++i) {
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int v = s[i - 1] - '0';
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st ^= 1 << v;
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if (~d[st]) {
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ans = max(ans, i - d[st]);
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} else {
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d[st] = i;
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}
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for (v = 0; v < 10; ++v) {
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if (~d[st ^ (1 << v)]) {
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ans = max(ans, i - d[st ^ (1 << v)]);
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}
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}
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}
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return ans;
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}
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};
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func longestAwesome(s string) int {
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d := [1024]int{}
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d[0] = 1
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st, ans := 0, 1
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for i, c := range s {
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i += 2
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st ^= 1 << (c - '0')
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if d[st] > 0 {
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ans = max(ans, i-d[st])
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} else {
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d[st] = i
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}
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for v := 0; v < 10; v++ {
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if d[st^(1<<v)] > 0 {
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ans = max(ans, i-d[st^(1<<v)])
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}
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}
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}
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return ans
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,23 @@
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class Solution {
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public int longestAwesome(String s) {
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int[] d = new int[1024];
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int st = 0, ans = 1;
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Arrays.fill(d, -1);
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d[0] = 0;
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for (int i = 1; i <= s.length(); ++i) {
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int v = s.charAt(i - 1) - '0';
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st ^= 1 << v;
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if (d[st] >= 0) {
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ans = Math.max(ans, i - d[st]);
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} else {
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d[st] = i;
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}
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for (v = 0; v < 10; ++v) {
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if (d[st ^ (1 << v)] >= 0) {
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ans = Math.max(ans, i - d[st ^ (1 << v)]);
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}
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}
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}
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return ans;
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}
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}
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,16 @@
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class Solution:
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def longestAwesome(self, s: str) -> int:
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st = 0
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d = {0: -1}
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ans = 1
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for i, c in enumerate(s):
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v = int(c)
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st ^= 1 << v
9+
if st in d:
10+
ans = max(ans, i - d[st])
11+
else:
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d[st] = i
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for v in range(10):
14+
if st ^ (1 << v) in d:
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ans = max(ans, i - d[st ^ (1 << v)])
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return ans

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