[LeetCode] Palindrome Partitioning II

原题地址：https://oj.leetcode.com/problems/palindrome-partitioning-ii/

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.

思路还是采用动态规划，和WordBreak很像，只不过dict需要事先自己算出，LeetCode的discussion里有一个方法可以边动规边计算dict，本质上没有区别，两次循环复杂度为O(n).

public class Solution {
public int minCut(String s) {
if (s == null || s.length() < 1) {
return 0;
}

int len = s.length();
// pal[i][j] == true, means s.substring(i, j + 1) is palindrome
// obsviously, when j - i < 1, pal[i][j] = true
boolean pal[][] = getDict(s);
// dp[i] is the min cuts of s.substring(i, len)
int dp[] = new int[len];
dp[len - 1] = 0;
for (int i = len - 2; i >= 0; i--) {
dp[i] = len - i - 1; // max cut
for (int j = i; j < len; j++) {
if (pal[i][j]) {
if (j == len - 1) {
dp[i] = 0;
break;
} else if (dp[j + 1] + 1 < dp[i]){
dp[i] = dp[j + 1] + 1;
}
}
}
}

return dp[0];
}

private boolean[][] getDict(String s) {
int len = s.length();
boolean[][] pal = new boolean[len][len];

for (int i = len - 1; i >= 0; i--) {
for (int j = i; j < len; j++) {
if (s.charAt(i) == s.charAt(j) && (j - i < 2 || pal[i + 1][j - 1])) {
pal[i][j] = true;
}
}
}

return pal;
}
}