LeetCode Palindrome Partitioning II

原题链接在这里：https://leetcode.com/problems/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.

题解：

与Word Break相似. DP问题, 求的是至少切几刀能全是palindrome. 需要保留的历史信息是到当前点最少能分成几块palindrome, 用array保存.

更新时如果[i,j]段是palindrome, 到j点结束那段就可以从到i点结束那段+1 来跟新最小值.

初始化至少每个字母都分开肯定都是palindrome了.

答案dp[len]-1. 最少分成的块数比切的刀数多一.

用二维array来村[i, j]段是否是palindrome. 两边闭区间，包括i,j对应的char.

Time Complexity: O(len^2). Space: O(len^2).

AC Java:

 1 public class Solution {
 2     public int minCut(String s) {
 3         if(s == null || s.length() == 0){
 4             return 0;
 5         }
 6         int len = s.length();
 7         boolean[][] isDic = helper(s);
 8         int [] res = new int[len+1];
 9         res[0] = 0;
10         for(int i = 0; i < len; i++){
11             res[i+1] = i+1;
12             for(int j = 0; j <= i; j++){        //error
13                 if(isDic[j][i]){
14                     res[i+1] = Math.min(res[i+1],res[j]+1);
15                 }
16             }
17         }
18         return res[len] - 1;
19     }
20
21     //helper function builds dictionary
22     private boolean[][] helper(String s){
23         int len = s.length();
24         boolean[][] dict = new boolean[len][len];
25         for(int i = len-1; i>=0; i--){
26             for(int j = i; j < len; j++){
27                 if(s.charAt(i) == s.charAt(j) && ((j-i)<2 || dict[i+1][j-1] )){     //error
28                     dict[i][j] = true;
29                 }
30             }
31         }
32         return dict;
33     }
34 }

可以省略掉dic. 对每一个点按照奇数和偶数两种方式左右延展若是palindrome就更新dp array. 若不是就停止.

Time Complexity: O(len^2). Space: O(len).

AC Java:

 1 class Solution {
 2     public int minCut(String s) {
 3         if(s == null || s.length() == 0){
 4             return 0;
 5         }
 6
 7         int len = s.length();
 8         int [] dp = new int[len+1];
 9         for(int i = 0; i<=len; i++){
10             dp[i] = i-1;
11         }
12         for(int i = 0; i<len; i++){
13             // odd length palindrome
14             for(int l = i, r = i; l>=0 && r<len && s.charAt(l)==s.charAt(r); l--, r++){
15                 dp[r+1] = Math.min(dp[r+1], dp[l]+1);
16             }
17
18             // even length palindrome
19             for(int l = i, r = i+1; l>=0 && r<len && s.charAt(l)==s.charAt(r); l--, r++){
20                 dp[r+1] = Math.min(dp[r+1], dp[l]+1);
21             }
22         }
23         return dp[len];
24     }
25 }

转载于:https://www.cnblogs.com/Dylan-Java-NYC/p/4824961.html

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