96：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.

解析*：该题目的解法思路和 Palindrome Partitioning 几乎一模一样，在依据动态规划算法判断字符串 s 的各个子字符串 s[i, j] 是否为回文的同时，给出最小切割数，代码如下:

// 动态规划，时间复杂度 O(n^2)，空间复杂度 O(n^2)
// 不仅判断字符串s 的各个子字符串s[i, j] 是否为回文,
// 同时在此过程中还求出字符串 s 的回文最小切割数
// 代码更为紧凑，简洁
class Solution {
public:
        int minCut(string s) {
                const int n = s.size();
                bool table[n][n] = {true}; // 判断 s[i,j] 是否为回文

                // min[i] 表示子字符串s[i, n)的最小回文切割数
                int min[n + 1];
                for (int i = 0; i <= n; ++i)
                        min[i] = n - i - 1; // 最后一个 min[n] = -1;

                for (int i = n - 1; i >= 0; --i) {
                        for (int j = i; j < n; ++j) {
                                int len = j - i + 1;
                                if (len == 1) table[i][j] = true;
                                else if (len = 2) table[i][j] = s[i] == s[j];
                                else table[i][j] = table[i + 1][j - 1] && s[i] == s[j];
                                min[i] = table[i][j] ? min(min[i], min[j + 1] + 1) : min[i];
                        }
                }
                return min[0];
        }
};