題目: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.
可以用DP的思路解這道題。
定義二維數組dp[i][j]記錄從i到j的子串是否是迴文;一維數組cutNum[i]記錄從i開始到字符串結尾的子串的最小分割次數。先將cotNum初始化爲每個子串的最大分割次數s.size() - i;(實際次數爲cutNum[i] - 1)
動規過程中的動態轉移方程是:cutNum[i] = min(cutNum[i],cutNum[j + 1] + 1)前提是子串i到j是迴文。
AC代碼:
class Solution {
public:
int minCut(string s) {
int len = s.size();
//dp二維數組記錄字符串從i到j是否是迴文
vector<vector<bool> > dp(len,vector<bool>(len,false));
//cutNum[i]記錄從i開始到字符串結尾的子串的最小分割次數
vector<int> cutNum(len + 1,0);
for(int i = len - 1;i >= 0;i--){
//先將最小切割次數初始化爲len - i
cutNum[i] = len - i;
for(int j = i;j < len;j++){
//如果從i到j的子串是迴文則s[i]肯定等於s[j]
if(s[i] == s[j]){
//s[i] == s[j]的情況下,如果j - i < 2或者從i+1到j-1的子串是迴文則從i到j的子串是迴文
if(j - i < 2 || dp[i + 1][j - 1]){
dp[i][j] = true;
//則從i開始到字符串結尾的子串用兩種切割情況:
//切割出從i到j的子串或者不切割,最小切割次數取其中比較小的值
cutNum[i] = cutNum[i] < (cutNum[j + 1] + 1)?cutNum[i] : (cutNum[j + 1] + 1);
}
}
}
}
return cutNum[0] - 1;
}
};