Palindrome Partitioning II 最小的括号分割

来源：互联网发布：淘宝运营pdf 编辑：程序博客网时间：2024/05/17 03:25

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.

前面一道题目是要找到所有的回文子字符组合，现在是找到最小的分割数。很显然，应该用动态规划优化。

开一个数组minValue， minValue[ i ] 存储以 index i 为起点的字符串的最小分割数。

同时开一个二维数组isPal, 用来存储string中子字符串是否为回文字符。( isPal [ i ] [ j ] = isPal [ i ] [ j ] && s [ i ] == s [ j ] )

运行时间：

代码：

public class PalindromePartitioningII {    public int minCut(String s) {        int[] minCache = new int[s.length()];        boolean[][] isPal = new boolean[s.length()][s.length()];        for (int i = s.length() - 1; i >= 0; i--) {            for (int j = i; j < s.length(); j++) {                if (( i + 1 > j - 1 || isPal[i + 1][j - 1]) && s.charAt(i)== s.charAt(j))                    isPal[i][j] = true;            }        }        Arrays.fill(minCache, Integer.MAX_VALUE);        minCache[s.length() - 1] = 0;        return doPartition(minCache, isPal, s, 0);    }    private int doPartition(int[] minCache, boolean[][] isPal, String s, int begin) {        if (begin == s.length()) {            return 0;        }        if (minCache[begin] != Integer.MAX_VALUE) {            return minCache[begin];        }        int curMin = Integer.MAX_VALUE;        for (int i = begin; i < s.length(); i++) {            if (isPal[begin][i]) {                if (i == s.length() - 1) {                    curMin = 0;                    break;                }                curMin = Math.min(curMin, doPartition(minCache, isPal, s, i + 1) + 1);            }        }        minCache[begin] = curMin;        return minCache[begin];    }}

1 0