【LeetCode】121.Palindrome Partitioning II
阿新 • • 發佈:2018-11-01
題目描述(Hard)
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.
題目連結
https://leetcode.com/problems/palindrome-partitioning-ii/description/
Example 1:
Input: "aab"
Output: 1
Explanation: The palindrome partitioning ["aa","b"] could be produced using 1 cut.
演算法分析
每次從i往右掃描,每找到一個迴文就算一次DP,可以轉換為f(i)=[i, n-1]之間的最小的cut數,n為字串長度,則狀態轉移方程為。判斷[i,j]為迴文,每次從i到j比較太過於費時,可以定義狀態P[i][j]=true,如果[i,j]為迴文,那麼P[i][j]=str[i]==str[j] && P[i+1][j-1]。
提交程式碼:
class Solution { public: int minCut(string s) { const int n = s.size(); int f[n + 1]; bool p[n][n]; fill_n(&p[0][0], n * n, false); for (int i = 0; i <= n; ++i) f[i] = n - i - 1; for (int i = n - 1; i >= 0; --i) { for (int j = i; j < n; ++j) { if (s[i] == s[j] && (j - i < 2 || p[i + 1][j - 1])) { p[i][j] = true; f[i] = min(f[i], f[j + 1] + 1); } } } return f[0]; } };