given a word,convert it into a pallindrome with minimum addition of letters to it.letters can be added anywhere in the word.for eg if yahoo is given result shud be yahoohay.give a optimize soln
方法1:
设字符串str1, 其倒置字符串str2, 求得str1和str2的最长公共序列str3
对str1,str2,str3做三路归并
方法2:
dp[i][dis]表示从i开始的长度为dis的字符串,要变为回文,最少添加多少字符
#include <iostream> #include <stdio.h> using namespace std; const int N = 1000; char ans[N]; int dp[N][N]; char s[N]; int tracebace(int p, int dis, int i, bool &flag) { if (dis == 0) { flag = false; return 0; } if (dis == 1) { flag = true; ans[i] = s[p]; return 1; } if (s[p] == s[p+dis-1]) { ans[i] = s[p]; return 1 + tracebace(p + 1, dis -2, i +1, flag); } if (dp[p][dis] == dp[p+1][dis-1] + 1) { ans[i] = s[p]; return 1 + tracebace(p + 1, dis - 1, i + 1, flag); } else { ans[i] = s[p+dis-1]; return 1 + tracebace(p, dis - 1, i + 1, flag); } } int main() { scanf("%s", s); memset(dp,0,sizeof(dp)); int len = strlen(s); for (int dis = 2; dis <= len; ++dis ) { for (int i = 0; i <= len - dis; ++i) { if (s[i] == s[i + dis - 1]) dp[i][dis] = dp[i + 1][dis - 2]; else { dp[i][dis] = 1 + min(dp[i+1][dis-1], dp[i][dis-1]); } } } bool flag = true; int preLen = tracebace(0, len, 0, flag); int dis = flag ? 1 : 0; for (int i = preLen; i < len + dp[0][len]; ++i,++dis) { ans[i] = ans[preLen-dis-1]; } ans[len + dp[0][len]] = 0; cout << ans<<endl; return 0; }