[leetcode] 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.

https://oj.leetcode.com/problems/palindrome-partitioning-ii/

思路：DP。cuts[i]表示从i到结尾的子串，要达到题意需要的最少切割数。isPalin用来判断是否是palindrome。

　　初始化：cuts[i]=len-i

　　推倒：cuts[i]=true if s.charAt(i) == s.charAt(j) && (j - i < 2 || isPalin[i + 1][j - 1])

　　推倒的同时求出isPalin数组的值，提高效率。

public class Solution {
public int minCut(String s) {
if (s == null || s.length() == 0)
return 0;
int len = s.length();
boolean[][] isPalin = new boolean[len][len];
int[] cuts = new int[len + 1];
for (int i = 0; i < len; i++)
cuts[i] = len - i;

for (int i = len - 1; i >= 0; i--) {
for (int j = i; j < len; j++) {
if (s.charAt(i) == s.charAt(j) && (j - i < 2 || isPalin[i + 1][j - 1])) {
isPalin[i][j] = true;
cuts[i] = Math.min(cuts[i], cuts[j + 1] + 1);
}
}

}

return cuts[0] - 1;

}

public static void main(String[] args) {
System.out.println(new Solution().minCut("bb"));
}
}

第二遍记录：注意DP的出事状态和递推关系，与第一遍解法不同，dp[i]代表s[0..i]的最小分割数，需要根据当前元素是否与之前元素组成回文不断更新最小值。

public int minCut(String s) {
if (s == null || s.length() == 0)
return 0;
int len = s.length();
boolean[][] isPalin = new boolean[len][len];
int[] cuts = new int[len + 1];
for (int i = 0; i <= len; i++)
cuts[i] = i;

for (int i = len - 1; i >= 0; i--) {
for (int j = i; j < len; j++) {
if (s.charAt(i) == s.charAt(j) && (j - i < 2 || isPalin[i + 1][j - 1])) {
isPalin[i][j] = true;
}
}

}

for (int i = 1; i <= len; i++) {for (int j = 1; j <= i; j++) {
if (isPalin[j - 1][i - 1]) {
cuts[i] = Math.min(cuts[i], cuts[j - 1] + 1);
}
}

}

return cuts[len] - 1;

}

第三遍记录：

　　dp[i]代表s[0..i]的最小分割数

　　注意dp数组初始状态

　　先用二位dp求出isPalin数组用于后面快速查询

/**
*
s: abaab
minCut:2

""  a    b    a    a    b    b
dp   -1  0    1    0    1    2    2
init -1  0    1    2    3    4    5

*
*/

public class Solution {
public int minCut(String s) {
if (s == null || s.length() <= 1)
return 0;
int n = s.length();
int[] dp = new int[n + 1];
for (int i = 0; i <= n; i++)
dp[i] = i - 1;

boolean[][] isPalin = new boolean

;
for (int i = n - 1; i >= 0; i--) {
for (int j = i; j < n; j++) {
if (s.charAt(i) == s.charAt(j) && (j - i < 2 || isPalin[i + 1][j - 1])) {
isPalin[i][j] = true;
}
}

}

for (int i = 1; i <= n; i++) {
for (int j = 1; j <= i; j++) {
if (isPalin[j - 1][i - 1]) {
dp[i] = Math.min(dp[j - 1] + 1, dp[i]);
}
}
}

return dp
;

}

public static void main(String[] args) {
System.out.println(new Solution().minCut("abababab"));
}
}

参考：

/article/1382266.html

http://www.tuicool.com/articles/jmQ3uu