LeetCode132：Palindrome Partitioning II

题目：

Given a string s, partition s such that every substring of the partition is a palindrome.

Return the minimum cuts needed for a palindrome partitioning of s.

For example, given s =

"aab"

,
Return

1

since the palindrome partitioning

["aa","b"]

could be produced using 1 cut.

解题思路：

这题是参考网上大牛的，一开始我采用上一题Palindrome Partitioning解题思路，结果运行超时。

实现代码：

#include <iostream>
#include <vector>
#include <iterator>
#include <unordered_map>
#include <string>
#include <algorithm>

using namespace std;
/*
Given a string s, partition s such that every substring of the partition is a palindrome.

Return the minimum cuts needed for a palindrome partitioning of s.

For example, given s = "aab",
Return 1 since the palindrome partitioning ["aa","b"] could be produced using 1 cut.
*/
class Solution {
public:
//DP
/*
如果if s[j...k] 为回文串，则 dp[k] = min{ dp[k], dp[j - 1] + 1}  0 <=j <= k - 1;
否则 dp[k] = dp[k - 1] + 1
*/
int minCut(string s){
if(s.size() == 0)
return -1;
//dp保存前i个字符串最小切割数
vector<int> dp(s.size() + 1, s.size()-1);//初始化最小切分数为s.size-1

//保存i—j是否为回文串
vector<vector<bool> > status(s.size(), vector<bool>(s.size(), false));
dp[0] = -1;
for(int i = 0; i < s.size(); ++i)
{

dp[i + 1] = dp[i] + 1;//假定，s[i]不能和其前面的字符串构成回文串，则s[i]为单独回文串即要切分
/*
检查s[i]能否和其前面的字符串构成回文串，如果可以，则更新dp[i+1]
*/
for(int cur = i - 1; cur >= 0; --cur)
if(s[i] == s[cur] && (i - cur <= 2 || status[cur + 1][i - 1])){
dp[i + 1] = min(dp[i + 1], dp[cur] + 1);
status[cur][i] = true;
}

}
return dp[s.size()];
}

};

int main(void)
{
string s("aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa");
//string s("aabbed");
Solution solution;
int minCut = solution.minCut(s);
cout<<minCut<<endl;
return 0;
}