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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.
本来是按照Palindrome Partitioning那题的思路改的，复杂度是O(n^3)吧，但是TLE了，考虑降低复杂度，算法中可以降低复杂度的只有判断字符串是否是回文那部分，因为每次计算没有考虑之前计算过的情况，将这部分提出来，用动态规划去做，isPa[i][j] = s.charAt(i) == s.charAt(j) && (i == j || i + 1 == j || isPa[i + 1][j - 1])，这样，判断回文的复杂度为O(n^2)，然后自底向上计算时复杂度为O(n^2)，总体的复杂度为O(n^2)。代码如下：
1 public int minCut(String s) { 2 boolean isPa[][] = new boolean[s.length()][s.length()]; 3 if (s == null || s.equals("")) { 4 return 0; 5 } 6 for (int i = s.length() - 1; i >= 0; i--) { 7 for (int j = i; j < s.length(); j++) { 8 isPa[i][j] = s.charAt(i) == s.charAt(j) 9 && (i == j || i + 1 == j || isPa[i + 1][j - 1]); 10 } 11 } 12 int result[] = new int[s.length()]; 13 for (int i = 0; i < s.length(); i++) { 14 result[i] = 100000; 15 } 16 for (int i = s.length() - 1; i >= 0; i--) { 17 for (int j = i; j < s.length(); j++) { 18 if (isPa[i][j]) { 19 if (j == s.length() - 1) { 20 result[i] = 0; 21 } else { 22 result[i] = result[i] < result[j + 1] + 1 ? result[i] 23 : result[j + 1] + 1; 24 } 25 } 26 } 27 } 28 return result[0]; 29 }
Palindrome Partitioning II,布布扣,bubuko.com
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原文地址：http://www.cnblogs.com/apoptoxin/p/3793158.html
