[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 + ];//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 = ; i <= len; ++i)
             dp[i] = len - i;
         //初始化palin[i][j]
         for(int i = ; i < len; ++i)
             for(int j = ; j < len; ++j)
                 palin[i][j] = false;
        //从后往前计算dp[i],
         for(int i = len - ; i >= ; --i)
             for(int j = i; j < len; ++j)
             {
                 if(s[i] == s[j] && (j - i <  || palin[i + ][j - ]))
                 {
                     palin[i][j] = true;
                     dp[i] = min(dp[i], dp[j + ] + );
                 }
             }
         return dp[] - ;
     }
 };