【leetcode】1312. Minimum Insertion Steps to Make a String Palindrome

题目如下：

Given a string s. In one step you can insert any character at any index of the string.

Return the minimum number of steps to make s palindrome.

A Palindrome String is one that reads the same backward as well as forward.

Example 1:

Input: s = "zzazz"
Output: 0
Explanation: The string "zzazz" is already palindrome we don't need any insertions.

Example 2:

Input: s = "mbadm"
Output: 2
Explanation: String can be "mbdadbm" or "mdbabdm".

Example 3:

Input: s = "leetcode"
Output: 5
Explanation: Inserting 5 characters the string becomes "leetcodocteel".

Example 4:

Input: s = "g"
Output: 0

Example 5:

Input: s = "no"
Output: 1 

Constraints:

• 1 <= s.length <= 500
• All characters of s are lower case English letters.

解题思路：动态规划。假设dp[i][i+j]为使得子串s[i:i+j]变成回文需要插入的最大字符数，那么有以下几种情况建立状态转移方程：

1，如果s[i] == s[i+j]，求下面三种情况的最小值

     a. s[i] 和 s[i+j] 配对，表示不需要额外插入字符，那么 dp[i][j] = dp[i+1][i+j-1]

     b. 在s[i+j]后面新插入一个字符和 s[i]配对， 那么 dp[i][j] = dp[i+1][i+j] + 1 

     c.在s[i]前面插入一个字符和 s[i+j]配对，那么 dp[i][j] = dp[i][i+j-1] + 1

2，如果s[i] != s[i+j]，求下面三种情况的最小值  

     a. 在s[i+j]后面新插入一个字符和 s[i]配对， 那么 dp[i][j] = dp[i+1][i+j] + 1 

     b. 在s[i]前面插入一个字符和 s[i+j]配对，那么 dp[i][j] = dp[i][i+j-1] + 1

     c. 在s[i+j]后面新插入一个字符和 s[i]配对，并且 在s[i]前面插入一个字符和 s[i+j]配对，那么 dp[i][j] = dp[i+1][i+j-1] + 2

最后dp[0][-1] 的值即为结果。

代码如下：

class Solution(object):
    def minInsertions(self, s):
        """
        :type s: str
        :rtype: int
        """
        dp = [[0] * len(s) for _ in s]
        for j in range(1,len(s)): #j : the lenght of substring
            for i in range(len(s)):
                if i + j >= len(s):continue
                elif i + 1 > i + j - 1 and s[i] == s[i+j]:
                    continue
                elif i + 1 > i + j - 1 and s[i] != s[i + j]:
                    dp[i][i + j] = 1
                elif s[i] == s[i+j]:
                    dp[i][i+j] = min(dp[i+1][i+j-1],dp[i+1][i+j]+1,dp[i][i+j-1] + 1)
                else:
                    dp[i][i + j] = min(dp[i + 1][i + j] + 1,dp[i + 1][i + j-1] + 2, dp[i][i + j-1] + 1)
        #print dp
        return dp[0][-1]