问题描述:
输入n行字符串,找出它们最后一个单词的最长公共尾串
方法:
可将第 i 个字符串反转过来,过滤先行空格,找到第一个单词(原字符串最后一个单词的反转),和第i + 1个字符串反转后的第一个单词匹配;
将匹配结果和第 i + 2 个字符串对比, etc.
代码:
#include <iostream> #include <string> #include <algorithm> using namespace std; string find_suffix(string& s1, string& s2) { // reverse(s1.begin(), s1.end()); reverse(s2.begin(), s2.end()); size_t s1_begin = 0; size_t s2_begin = 0; for ( ; s1[s1_begin] == ' '; ++ s1_begin) {} for ( ; s2[s2_begin] == ' '; ++ s2_begin) {} size_t s1_end = s1_begin; size_t s2_end = s2_begin; for ( ; s1_end < s1.size() && s2_end < s2.size() && s1[s1_end] == s2[s2_end]; ++ s1_end, ++ s2_end ) {} return s1.substr(s1_begin, s1_end-s1_begin); } int main() { int n; string s1, s2; cin >> n; getchar(); getline(cin ,s1); reverse(s1.begin(), s1.end()); for (int i = 2; i <= n; ++ i) { getline(cin, s2); s1 = find_suffix(s1, s2); } if (s1.size() != 0) { reverse(s1.begin(), s1.end()); cout << s1 << endl; } else { cout << "nai" << endl; } return 0; }