回文串的问题很经典，也很常见，涉及到递归，循环，动态规划等方面，这里总结一下几种类型，供以后回顾，有问题请大家指正
1、回文串的判断
leetcode上的题目
bool isPalindrome(const char* src)
{
if(src == NULL)return true;
int end = strlen(src)-1,begin = 0;
while(begin < end)//从两边向中间判断，当然也可以从中间向两边判断
{
if(src[begin] != src[end])return false;
begin ++;
end --;
}
return true;
}void LongestPalindromeDP(const char* src)//DP算法
{
if(src == NULL)return;
int length = strlen(src);
bool dp[length][length];
int i,j,len,maxLen = 1,begin = 0,end = 0;
for(i = 0;i < length;i++)
{
memset(dp[i],false,sizeof(bool)*length);
}
for(len = 2;len <= length;len++)
{
for(i = 0;i + len - 1 < length;i++)
{
j = i+len-1;
if(src[i] == src[j] && (j - i <= 2 || dp[i+1][j-1]))//dp[i][j]表示从第i到第j个元素是否为回文串
{
dp[i][j] = true;
begin = i;
end = j;
maxLen = len;
}
}
}
for(i = begin;i <= end;i++)cout << src[i];
cout << endl;
}
int expandAroundCenter(const char* src,int left,int right)//从中间往两头判断最长回文串
{
int length = strlen(src);
while(left >= 0 && right < length && src[left] == src[right])
{
left --;
right ++;
}
return right - left - 1;
}
void LongestPalindromeCenter(const char* src)
{
if(src == NULL)return;
int length = strlen(src);
int i,maxLen = 1,begin = 0,end = 0;
for(i = 0;i < length;i++)
{
int len = expandAroundCenter(src,i,i);//针对奇数
if(len > maxLen)
{
maxLen = len;
begin = i - ((len-1) >> 1);
end = i + ((len-1) >> 1);
}
len = expandAroundCenter(src,i,i+1);//针对偶数
if(len > maxLen)
{
maxLen = len;
begin = i - ((len-1) >> 1);
end = i + 1 + ((len-1) >> 1);
}
}
for(i = begin;i <= end;i++)cout<< src[i];
cout << endl;
}int changeToPalindrome(const char* src)
{
if(src == NULL)return 0;
int length = strlen(src);
int dp[length][length];
int i,j,len;
for(i = 0;i < length;i++)memset(dp[i],0,sizeof(int)*length);
for(len = 2;len <= length;len++)
{
for(i = 0;i + len - 1 < length;i++)
{
j = i + len - 1;
if(src[i] == src[j])dp[i][j] = dp[i+1][j-1];//dp[i][j]表示从第i个到第j个需要添加的字符数
else dp[i][j] = 1 + min(dp[i+1][j],dp[i][j-1]);//在前面或者后面添加一个字符
}
}
return dp[0][length-1];
}bool isPalindrome(int x) {
if(x < 0)return false;
int base = 1;
while(x / 10 >=base)base *= 10;//获取最高位的权值
while(x > 0)
{
int left = x / base;//最高位
int right = x % 10;//最低位
if( left != right)return false;
x -= base*left;
x /= 10;//除去最高最低位
base /= 100;
}
return true;
}附，把一个数字reverse
int reverseNum(int num)
{
int MAX = numeric_limits<int>::max();
int MIN = numeric_limits<int>::min();
int res = 0,flag = 1;
if(num < 0)flag = -1;
while(num != 0)
{
int low = num % 10;//商和余数的符合和被除数相同
if(flag == 1 && (MAX / 10 < res || (MAX / 10 == res && low > MAX % 10)))//溢出测试不能用乘法，要用除法
{
cout << "over flow 1"<<endl;
return -1;
}
else if(flag == -1 && (MIN / 10 > res || (MIN / 10 == res && low < MIN % 10)))
{
cout << "over flow 2" << endl;
return -1;
}
res = (res * 10)+low;
num /= 10;
}
return res;
}int minCut(string s) {
int length = s.size();
if(length <= 0)return 0;
bool isParlindrome[length][length];
int i,j;
for(i = 0;i < length;i++)
{
memset(isParlindrome[i],false,sizeof(bool)*length;
}
int dp[length+1];
for(i = length;i >= 0;i--)dp[i] = length - i - 1;
for(i = length-1;i >= 0 ;i--)
{
for(j = i;j < length;j++)
{
if(s[i] == s[j] && (j - i <= 2 || isParlindrome[i+1][j-1]))
{
isParlindrome[i][j] = true;
dp[i] = min(dp[i],1+dp[j+1]);
}
}
}
return dp[0];
}bool isPalindrome(string s,set<string>& hash)
{
set<string>::iterator iter = hash.find(s);
if(iter != hash.end())return true;//如果该部分已经判断过，则直接返回
int i = 0,j = s.size()-1;
while(i < j)
{
if(s[i++] != s[j--])return false;
}
hash.insert(s);
return true;
}
void dfs(string s,vector<vector<string> >& res,vector<string>& v,set<string>& hash)
{
int length = s.size();
if(length == 0)
{
res.push_back(v);
return;
}
int i;
for(i = 1;i <= length;i++)
{
string sub = s.substr(0,i);
if(isPalindrome(sub,hash))//首先判断前半部分是不是回文串
{
v.push_back(sub);
dfs(s.substr(i),res,v,hash);//递归调用后半部分
v.pop_back();
}
}
}
vector<vector<string> > partition(string s) {
int length = s.size();
vector<vector<string> > res;
vector<string> v;
set<string> hash;
if(length <= 0)return res;
dfs(s,res,v,hash);//递归判断
return res;
}原文地址：http://blog.csdn.net/fangjian1204/article/details/38084165
