Problem Description
Give you a string s, please split s into some substrings so that each substring is a palindrome.
Returns the minimum number of splits that meet the requirements.
Example 1:
输入：s = "aab"
输出：1
解释：只需一次分割就可将 s 分割成 ["aa","b"] 这样两个回文子串。
Example 2:
输入：s = "a"
输出：0
Example 3:
输入：s = "ab"
输出：1
prompt:
1 <= s.length <= 2000
s 仅由小写英文字母组成
Problem-solving ideas
Let f[i] denote the minimum number of divisions of the prefix s[0...i] of the string. To get the value of f[i], we can consider enumerating the last palindrome divided by s[0...i], so that we can write the state transition equation:
that is, we enumerate the last palindrome The starting position of j+1, to ensure that s[j+1...i] is a palindrome string, then f[i] can be transferred from f[j] with an additional number of divisions.
Note that in the above state transition equation, we have yet to consider a case, that is, s[0...i] itself is a palindrome string. At this time, it does not need to be divided, namely:
f[i]=0
So how do we know whether s[j+1...i] or s[0...i] is a palindrome? We can use the same pre-processing method as in Leetcode Problem Solution 5-Longest Palindrome Substring to calculate whether each substring of the string ss is a palindrome in advance, namely:
Code
class Solution {
public boolean f[][]; //判断以i为起点，j为终点的字符串是否是回文串
public int n; //字符串s的长度
public int dp[]; //dp[i]表示以i为最后一个元素的子字符串的最少分割次数
public int minCut(String s) {
n=s.length();
//按最长回文子串的思路填充f[][]
f=new boolean[n][n];
if(n==0 || n==1){
return 0;
}
char []cs=s.toCharArray();
for(int i=0;i<n;i++){
for(int j=0;j<=i;j++){
if(i-j<=1){
if(cs[i]==cs[j]){
f[j][i]=true;
}
}else{
if(cs[i]==cs[j] && f[j+1][i-1]==true){
f[j][i]=true;
}
}
}
}
dp=new int[n];
//计算以每个元素为后缀的子字符串的最少分割次数
for(int i=0;i<n;i++){
dp[i]=i; //最坏情况下，默认每个字符都要分割
//如果从第一个元素一直到最后一个元素都是回文子串，那么最少分割次数是0
if(f[0][i]==true){
dp[i]=0;
continue;
}
//否则的话，从前往后遍历，如果从j+1到i可以组成一个回文子串
//那么最少分割次数就等于0到j的最少分割次数+1
for(int j=0;j<i;j++){
if(f[j+1][i]==true){
dp[i]=Math.min(dp[j]+1,dp[i]);
}
}
}
//最终返回dp[n-1]
return dp[n-1];
}
}