【leetcode】 Palindrome Partitioning II

From： https://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.

倒着

class Solution {
public:
int minCut(string s) {
int len = s.size();
// i~len的最少的回文数，比cut数大1
vector<int> d(len+1);
// i与j是否相等
vector<vector<bool>> m(len, vector<bool>(len, false));
for(int i=len; i>=0; i--) d[i]=len-i;

for(int i=len-1; i>=0; i--) {
// 计算d[i]
for(int j=i; j<len; j++) {
//  d[i] = min(d[i], d[j+1]+1);
//  往中间挤
if(s[i]==s[j] && (j-i<2||m[i+1][j-1])) {
m[i][j] = true;
d[i] = min(d[i], d[j+1]+1);
}
}
}
return d[0]-1;
}
};

顺着

public class Solution {
public int minCut(String s) {
int l;
if (s == null || (l = s.length()) == 0) {
return 0;
}
// 从0~i（含i）的最小回文数
int[] palins = new int[l];
for (int i = 0; i < l; ++i) {
palins[i] = i + 1;
}
// sameAt[i][j]:i与j之间是回文
boolean[][] sameAt = new boolean[l][l];
for (int i = 0; i < l; ++i) {
// 计算palins[i]
for (int j = i; j >= 0; --j) {
if (s.charAt(i) == s.charAt(j) && (i - j < 2 || sameAt[i - 1][j + 1])) {
sameAt[i][j] = true;
int palin = (j > 0 ? palins[j - 1] : 0) + 1;
if (palin < palins[i]) {
palins[i] = palin;
}
}
}
}
return palins[l - 1] - 1;
}
}