标签：show splay als 字符串 ons code star 否则 for
// 面试题48：最长不含重复字符的子字符串 // 题目：请从字符串中找出一个最长的不包含重复字符的子字符串，计算该最长子 // 字符串的长度。假设字符串中只包含从‘a‘到‘z‘的字符。 #include <string> #include <iostream> // 方法一：蛮力法 bool hasDuplication(const std::string& str, int position[]); int longestSubstringWithoutDuplication_1(const std::string& str) { int length = str.length(); int longest = 0; int* position = new int[26]; //26个字母 for (int start = 0; start < length; ++start) { for (int end = start; end < length; ++end) { int count = end - start + 1; const std::string& substring = str.substr(start, count); //裁剪出子字符串 if (!hasDuplication(substring, position)) { if (count > longest) longest = count; } else break; } } delete[] position; return longest; } bool hasDuplication(const std::string& str, int position[]) { for (int i = 0; i < 26; ++i) position[i] = -1; for (int i = 0; i < str.length(); ++i) { int indexInPosition = str[i] - ‘a‘; if (position[indexInPosition] >= 0) //字符串出现过2次以上 return true; position[indexInPosition] = indexInPosition; } return false; } // 方法二：动态规划 int longestSubstringWithoutDuplication_2(const std::string& str) { int maxLength = 0; int curLength = 0; //每个字母上一次出现的位置 int* position = new int[26]; for (int i = 0; i < 26; ++i) position[i] = -1; for (int i = 0; i < str.length(); ++i) { int indexPrev = position[str[i] - ‘a‘]; //该字母上次出现的位置 if (indexPrev < 0 || i - indexPrev > curLength) //当前字母未出现过, 或上次出现在当前序列外 ++curLength; else { if (curLength > maxLength) maxLength = curLength; curLength = i - indexPrev; //否则以当前字母上次出现的位置后到当前字母位置作为当前长度 } position[str[i] - ‘a‘] = i; //更新位置索引 } if (curLength > maxLength) maxLength = curLength; delete[] position; return maxLength; }
// ====================测试代码==================== void testSolution1(const std::string& input, int expected) { int output = longestSubstringWithoutDuplication_1(input); if (output == expected) std::cout << "Solution 1 passed, with input: " << input << std::endl; else std::cout << "Solution 1 FAILED, with input: " << input << std::endl; } void testSolution2(const std::string& input, int expected) { int output = longestSubstringWithoutDuplication_2(input); if (output == expected) std::cout << "Solution 2 passed, with input: " << input << std::endl; else std::cout << "Solution 2 FAILED, with input: " << input << std::endl; } void test(const std::string& input, int expected) { testSolution1(input, expected); testSolution2(input, expected); } void test1() { const std::string input = "abcacfrar"; int expected = 4; test(input, expected); } void test2() { const std::string input = "acfrarabc"; int expected = 4; test(input, expected); } void test3() { const std::string input = "arabcacfr"; int expected = 4; test(input, expected); } void test4() { const std::string input = "aaaa"; int expected = 1; test(input, expected); } void test5() { const std::string input = "abcdefg"; int expected = 7; test(input, expected); } void test6() { const std::string input = "aaabbbccc"; int expected = 2; test(input, expected); } void test7() { const std::string input = "abcdcba"; int expected = 4; test(input, expected); } void test8() { const std::string input = "abcdaef"; int expected = 6; test(input, expected); } void test9() { const std::string input = "a"; int expected = 1; test(input, expected); } void test10() { const std::string input = ""; int expected = 0; test(input, expected); } int main(int argc, char* argv[]) { test1(); test2(); test3(); test4(); test5(); test6(); test7(); test8(); test9(); test10(); return 0; }
分析：动态规划，递归。
标签：show splay als 字符串 ons code star 否则 for
原文地址：https://www.cnblogs.com/ZSY-blog/p/12636233.html
