108

2017.10.24

理论上应该是很简单的   d[i][j] = min(d[i][k]+d[k+1][j]+1,d[i][j],但是如果单纯使用递归的话，显然会超时的。

后来采用数组记录d[i][j]的值，基本通过97%，但仍旧会超时。

在后来改变判断回文的办法，同样采用数组进行保存，这样就过了。

public class Solution {    /**     * @param s a string     * @return an integer     */public int minCut(String s) {        // write your code hereboolean[][] isHui = new boolean[s.length()][s.length()];isHui = isHuiWen(isHui,s);if(s.length() <= 1){return 0;}//建立数组，记录 arr[i][j]表示的是从第i个字符到第j个字符的最小分割数int [][]arr = new int[s.length()][s.length()];for(int i = 0; i < s.length()-1; i++){if(isHui[i][i+1] == true){arr[i][i+1] = 0;}else{arr[i][i+1] = 1;}}for(int i = 2; i < s.length(); i++){for(int j = 0; j < s.length()- i; j++){arr[j][j+i] = i;if(isHui[j][j+i] == true){arr[j][j+i] = 0;}else{for(int k = 0; k < i; k ++){if(arr[j][j+i] == 1){break;}arr[j][j+i] = Math.min(arr[j][j+k] + arr[j+k+1][j+i] + 1,arr[j][j+i]);}}}}return arr[0][s.length()-1];    }public boolean[][] isHuiWen(boolean[][] isHui,String s){for(int i = 0; i < s.length(); i++){isHui[i][i] = true;if(i < s.length()-1 && s.charAt(i) == s.charAt(i+1)){isHui[i][i+1] = true;}}for(int i = 2; i < s.length(); i++){for(int j = 0; j < s.length()- i; j++){if(s.charAt(j) != s.charAt(j+i)){isHui[j][i+j] = false;continue;}isHui[j][i+j] = isHui[j+1][i+j-1];}}return isHui;}};