[LeetCode]切割字符串，使各个子串都是回文

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.

思路：

用数组dp[i]表示从下标i开始的字符串分割为回文子串的最少个数，

paliin[i][j] 表示从下标i到下标j是否为回文串，

利用动态规划

dp[i] = min(dp[i], 1 + dp[j])  if  (palin[i + 1][j - 1] == ture or j - i < 2 ) && s[i]  = s[j]   , i <= j < len , 

i从len - 1递减至0

总的时间复杂度为O(n^2)

code

class Solution {
public:
    int minCut(string s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int len = s.size();
        int dp[len + 1];//记录从i开始的最短切割次数
        bool palin[len][len];//记录palin(i,j)是否为回文
        //初始化dp[i]
        for(int i = 0; i <= len; ++i)
            dp[i] = len - i;
        //初始化palin[i][j]
        for(int i = 0; i < len; ++i)
            for(int j = 0; j < len; ++j)
                palin[i][j] = false;
       //从后往前计算dp[i],
        for(int i = len - 1; i >= 0; --i)
            for(int j = i; j < len; ++j)
            {
                if(s[i] == s[j] && (j - i < 2 || palin[i + 1][j - 1]))
                {
                    palin[i][j] = true;
                    dp[i] = min(dp[i], dp[j + 1] + 1);
                }
            }
        return dp[0] - 1;
    }
};