转换到回文字符串

来源：互联网发布：手机表格制作软件编辑：程序博客网时间：2024/06/07 06:10

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;}