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.
这个两个Dp， 一个是对那一部分是palindrome的二维dp。 还有一个是 对cut个数的一维dp。
public class Solution {
int[][] map = null;
public int minCut(String s) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
map = new int[s.length()][s.length()];
int[] cut = new int[s.length()];
for(int i = 0; i < s.length(); i++)
map[i][i] = 1;
for(int i = 0; i < s.length() - 1; i ++){
if(s.charAt(i) == s.charAt(i + 1)) map[i][i + 1] = 1;
else map[i][i + 1] = -1;
}
for(int i = 0; i < s.length(); i ++){
cut[i] = i;
for(int j = i; j < s.length(); j ++){
map[i][j] = checkPartition(s, i, j);
}
}
for(int j = 0; j < s.length(); j ++){
for(int i = 0; i < j; i ++){
if(map[i][j] == 1) cut[j] = Math.min(cut[j], 1 + cut[i]);
}
}
return cut[s.length() - 1];
}
public int checkPartition(String s, int start, int end){
if(map[start][end] != 0) return map[start][end];
if(s.charAt(start) != s.charAt(end)) return -1;
return checkPartition(s, start + 1, end - 1);
}
}
public class Solution {
public int minCut(String s) {
int leng = s.length();
if(leng == 0 || leng == 1) return 0;
boolean[][] isPal = new boolean[leng][leng];
int[] dp = new int[leng];
for (int i = 0; i < leng; i++) {
dp[i] = leng - 1 - i;
}
for (int i = leng - 1; i >= 0; --i) {
for (int j = i; j < leng; ++j) {
if (s.charAt(i) == s.charAt(j) && (j <= i + 2 || (i + 1 < leng && j - 1 >= 0 && isPal[i + 1][j - 1]))) {
isPal[i][j] = true;
if(j+1 < leng){
dp[i] = Math.min(dp[i], 1 + dp[j + 1]);
}else {
dp[i] = 0;
}
}
}
}
return dp[0];
}
}