[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;    }}