LeetCode OJ Palindrome Partitioning II
2015-03-22 18:34
337 查看
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 =
Return
be produced using 1 cut.
还想按照上一题的dfs解法来解,结果超时了,后来在discuss里面看到了更好地解法,用dp:
https://oj.leetcode.com/discuss/6691/my-dp-solution-explanation-and-code
直接摘录了:
Calculate and maintain 2 DP states:
pal[i][j] , which is whether s[i..j] forms a pal
d[i], which is the minCut for s[i..n-1]
Once we comes to a pal[i][j]==true:
if j==n-1, the string s[i..n-1] is a Pal, minCut is 0, d[i]=0;
else: the current cut num (first cut s[i..j] and then cut the rest s[j+1...n-1]) is 1+d[j+1], compare it to the exisiting minCut num d[i], repalce if smaller.
d[0] is the answer.
代码也是照抄,罪过:
Return the minimum cuts needed for a palindrome partitioning of s.
For example, given s =
"aab",
Return
1since the palindrome partitioning
["aa","b"]could
be produced using 1 cut.
还想按照上一题的dfs解法来解,结果超时了,后来在discuss里面看到了更好地解法,用dp:
https://oj.leetcode.com/discuss/6691/my-dp-solution-explanation-and-code
直接摘录了:
Calculate and maintain 2 DP states:
pal[i][j] , which is whether s[i..j] forms a pal
d[i], which is the minCut for s[i..n-1]
Once we comes to a pal[i][j]==true:
if j==n-1, the string s[i..n-1] is a Pal, minCut is 0, d[i]=0;
else: the current cut num (first cut s[i..j] and then cut the rest s[j+1...n-1]) is 1+d[j+1], compare it to the exisiting minCut num d[i], repalce if smaller.
d[0] is the answer.
代码也是照抄,罪过:
class Solution { public: int minCut(string s) { if(s.empty()) return 0; int n = s.size(); vector<vector<bool>> pal(n,vector<bool>(n,false)); vector<int> d(n); for(int i=n-1;i>=0;i--) { d[i]=n-i-1; // initialize the d[i] as its max(to a n-length string, the max cut is n - 1) for(int j=i;j<n;j++) { if(s[i]==s[j] && (j-i<2 || pal[i+1][j-1])) // if a palindrome's two sides are the same of its length is zero { pal[i][j]=true; if(j==n-1) d[i]=0; else if(d[j+1]+1<d[i]) d[i]=d[j+1]+1; } } } return d[0]; } };
相关文章推荐
- LeetCode 132. Palindrome Partitioning II(回文切分)
- [leetcode] Palindrome Partitioning II
- [LeetCode] Palindrome Partitioning II 解题笔记
- LeetCode Palindrome Partitioning II
- Leetcode 132. Palindrome Partitioning II
- Leetcode--Palindrome Partitioning II
- [LeetCode] Palindrome Partitioning II
- [leetcode]Palindrome Partitioning II
- 【LeetCode】Palindrome Partitioning II 解题报告
- LeetCode 132 Palindrome Partitioning II
- [leetcode]Palindrome Partitioning II
- leetcode之 Palindrome Partitioning I&II
- 【leetcode】Palindrome Partitioning && Palindrome Partitioning II
- 【leetcode刷题笔记】Palindrome Partitioning II
- leetCode解题报告之Palindrome Partitioning I,II(DFS,DP)
- Java for LeetCode 132 Palindrome Partitioning II
- [leetcode-132]Palindrome Partitioning II(java)
- [LeetCode]题解(python):132-Palindrome Partitioning II
- Leetcode: palindrome-partitioning && palindrome-partitioning-ii
- LeetCode – Refresh – Palindrome Partitioning II