题目

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.

方法

使用DP思想

先求解 i-j的字串是不是回文串，并将结果放入一个二维数组中去。

创建一个长度为字符串长度为n的数组dis，dis[k]:表示0-k字串的最小划分。

从0到n-1计算dis数组。

	boolean[][] palindromeArray(String s) {
		int len = s.length();
		boolean[][] arrayPal = new boolean[len][len];
		for (int interval = 0; interval < len; interval++) {
			for (int i = 0; i < len - interval; i++) {
				int j = i + interval;
				if (i == j){
					arrayPal[i][j] = true;
				} else if (s.charAt(i) == s.charAt(j)) {
					if (j == i + 1) {
						arrayPal[i][j] = true;
					} else if (arrayPal[i + 1][j - 1] == true) {
						arrayPal[i][j] = true;
					} else {
						arrayPal[i][j] = false;
					}
				} else {
					arrayPal[i][j] = false;
				}
			}
		}		
		return arrayPal;
	}
    public int minCut(String s) {
    	if (s == null) {
    		return -1;
    	} else if (s.length() == 0){
    		return 0;
    	} else if (s.length() == 1) {
    		return 0;
    	} else {
        	int len = s.length();
        	boolean[][] status = palindromeArray(s);
        	
        	int[] dis = new int[len];
        	for (int i = 0; i < len; i++) {
        		dis[i] = i;
        	}
        	for (int i = 1; i < len; i++) {
        		if (status[0][i] == true) {
        			dis[i] = 0;
        		} else {
            		for (int j = i; j > 0; j--) {
            			if (status[j][i] == true) {
            				int temp = dis[j - 1] + 1;
            				if (temp < dis[i]) {
            					dis[i] = temp;
            				}
            			}
            		}
        		}

        	}  
        	return dis[len - 1];      	        	
    	}
    }