【LeetCode】132.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.

Example 1:

Input: s = "aab"
Output: 1
Explanation: The palindrome partitioning ["aa","b"] could be produced using 1 cut.

Example 2:

Input: s = "a"
Output: 0

Example 3:

Input: s = "ab"
Output: 1

Constraints:

• 1 <= s.length <= 2000
• s consists of lower-case English letters only.

这种回文什么的，数据范围又小，一看就是 dp。但判断是否是回文数会很耗时间，因此我们先做一下预处理。

对于位置 i，它自身是一个回文串。然后我们向两边扩展，如果 s[i-j]==s[i+j]，那么从 i-j 到 i+j 就是个回文串。一旦出现 s[i-j]!=s[i+j]，那么剩下的就一定都不是回文串了。

上面是回文串元素个数为奇数的情况，偶数的同理，判断 s[i-j]==s[i+j+1] 即可。这样我们就得到了任意字符串是否为回文串，时间复杂度 O ( n 2 ) O(n^2) O(n2)

然后就是 dp，定义 f[i] 为从 0 到 i 的子串最少要分割多少次，状态转移方程为 f[i]=min(f[i],f[j-1]+1)，其中，j 到 i 的子串为回文串。初始条件为 f[0]=0。时间复杂度 O ( n 2 ) O(n^2) O(n2)

下面是代码

class Solution {
public:
    int minCut(string s) {
        bool isPartition[2001][2001];//判断是否为回文串
        int f[2001];
    	int len = s.length();
    	for (int i = 0; i < len; i++) {//回文串元素数目为奇数
        	isPartition[i][i] = true;
        	int j = 1;
        	while (i - j >= 0 && i + j < len && s[i - j] == s[i + j]) {
            	isPartition[i - j][i + j] = true;
            	j++;
        	}
        	while (i - j >= 0 && i + j < len) {
            	isPartition[i - j][i + j] = false;
            	j++;
        	}
    	}
    	for (int i = 0; i < len - 1; i++) {//回文串元素数目为偶数
        	int j = 0;
        	while (i - j >= 0 && i + j + 1 < len && s[i - j] == s[i + j + 1]) {
            	isPartition[i - j][i + j + 1] = true;
            	j++;
        	}
        	while (i - j >= 0 && i + j + 1 < len) {
            	isPartition[i - j][i + j + 1] = false;
            	j++;
        	}
    	}
    	f[0]=0;
    	for (int i = 1; i < len; i++) {//dp
        	f[i] = 9999;//取一个较大的数
        	if (isPartition[0][i])
            	f[i] = 0;
        	else
            	for (int j = 1; j <= i; j++)
                	if (isPartition[j][i])
                    	f[i] = min(f[i], f[j - 1] + 1);
    	}
    	return f[len - 1];
    }
};

实际上做一做剪枝还可以更快，比如判断回文的同时就进行 dp，把最小分割次数给算出来。状态转移方程 f[i+j+1] = min(f[i+j+1],1+f[i-j]) 和 f[i+j+1] = min(f[i+j+1],1+f[i-j+1])

class Solution {
public:
    int minCut(string s) {
        int len = s.length();
        int f[len + 1];
        for (int i = 0; i <= len; i++)
            f[i] = i-1;
        for (int i = 0; i < len; i++) {
            for (int j = 0; i-j >= 0 && i+j < len && s[i-j]==s[i+j] ; j++)
                f[i+j+1] = min(f[i+j+1],1+f[i-j]);
            for (int j = 1; i-j+1 >= 0 && i+j < len && s[i-j+1] == s[i+j]; j++)
                f[i+j+1] = min(f[i+j+1],1+f[i-j+1]);
        }
        return f[len];
    }
};

这下子终于全部进了前 1%

Runtime: 4 ms, faster than 99.85% of C++ online submissions for Palindrome Partitioning II.

Memory Usage: 6.1 MB, less than 99.88% of C++ online submissions for Palindrome Partitioning II.

标签：Partitioning,isPartition,int,++,len,II,132,&&,回文
来源： https://blog.csdn.net/qq_39678161/article/details/119908709